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Checking Invertibility of Square Matrix
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
The question is:
Which statement is the same as asserting a square matrix $ZinmathbbR^(n,n)$ is invertible.
- That the columns of $Z$ span $mathbbR^n$.
- That $Z$ has no eigenvectors.
- That $Z$ has no eigenvectors of eigenvalue $0$.
- $mathrmkerZ = mathbf0$
D is true from the rank-nullity theorem and properties of linear maps.
C is true since $0$ as an eigenvalue is equivalent to noninvertible.
B is false by extension of C
I am unsure of A.
linear-algebra matrices linear-transformations
edited Aug 3 at 13:36
zzuussee
1,101419
asked Aug 3 at 13:15
PERTURBATIONFLOW
396
add a comment |Â
up vote
1
down vote
favorite
The question is:
Which statement is the same as asserting a square matrix $ZinmathbbR^(n,n)$ is invertible.
- That the columns of $Z$ span $mathbbR^n$.
- That $Z$ has no eigenvectors.
- That $Z$ has no eigenvectors of eigenvalue $0$.
- $mathrmkerZ = mathbf0$
D is true from the rank-nullity theorem and properties of linear maps.
C is true since $0$ as an eigenvalue is equivalent to noninvertible.
B is false by extension of C
I am unsure of A.
linear-algebra matrices linear-transformations
edited Aug 3 at 13:36
zzuussee
1,101419
asked Aug 3 at 13:15
PERTURBATIONFLOW
396
please take the time to format properly the question using either mathjax or latex
– Ahmad Bazzi
Aug 3 at 13:18
@AhmadBazzi is there a tutorial on using mathjax or latex?
– PERTURBATIONFLOW
Aug 3 at 13:18
and also is it statement or statement(s) ?
– Ahmad Bazzi
Aug 3 at 13:21
@AhmadBazzi it's a group of individual statements ordered A, B, C, D from top to the bottom.
– PERTURBATIONFLOW
Aug 3 at 13:23
math formatting tutorial/reference
– Omnomnomnom
Aug 3 at 13:26
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The question is:
Which statement is the same as asserting a square matrix $ZinmathbbR^(n,n)$ is invertible.
- That the columns of $Z$ span $mathbbR^n$.
- That $Z$ has no eigenvectors.
- That $Z$ has no eigenvectors of eigenvalue $0$.
- $mathrmkerZ = mathbf0$
D is true from the rank-nullity theorem and properties of linear maps.
C is true since $0$ as an eigenvalue is equivalent to noninvertible.
B is false by extension of C
I am unsure of A.
linear-algebra matrices linear-transformations
edited Aug 3 at 13:36
zzuussee
1,101419
asked Aug 3 at 13:15
PERTURBATIONFLOW
396
The question is:
Which statement is the same as asserting a square matrix $ZinmathbbR^(n,n)$ is invertible.
- That the columns of $Z$ span $mathbbR^n$.
- That $Z$ has no eigenvectors.
- That $Z$ has no eigenvectors of eigenvalue $0$.
- $mathrmkerZ = mathbf0$
D is true from the rank-nullity theorem and properties of linear maps.
C is true since $0$ as an eigenvalue is equivalent to noninvertible.
B is false by extension of C
I am unsure of A.
linear-algebra matrices linear-transformations
edited Aug 3 at 13:36
zzuussee
1,101419
asked Aug 3 at 13:15
PERTURBATIONFLOW
396
edited Aug 3 at 13:36
zzuussee
1,101419
edited Aug 3 at 13:36
zzuussee
1,101419
edited Aug 3 at 13:36
zzuussee
1,101419
1,101419
asked Aug 3 at 13:15
PERTURBATIONFLOW
396
asked Aug 3 at 13:15
PERTURBATIONFLOW
396
asked Aug 3 at 13:15
PERTURBATIONFLOW
396
396
please take the time to format properly the question using either mathjax or latex
– Ahmad Bazzi
Aug 3 at 13:18
@AhmadBazzi is there a tutorial on using mathjax or latex?
– PERTURBATIONFLOW
Aug 3 at 13:18
and also is it statement or statement(s) ?
– Ahmad Bazzi
Aug 3 at 13:21
@AhmadBazzi it's a group of individual statements ordered A, B, C, D from top to the bottom.
– PERTURBATIONFLOW
Aug 3 at 13:23
math formatting tutorial/reference
– Omnomnomnom
Aug 3 at 13:26
add a comment |Â
please take the time to format properly the question using either mathjax or latex
– Ahmad Bazzi
Aug 3 at 13:18
@AhmadBazzi is there a tutorial on using mathjax or latex?
– PERTURBATIONFLOW
Aug 3 at 13:18
and also is it statement or statement(s) ?
– Ahmad Bazzi
Aug 3 at 13:21
@AhmadBazzi it's a group of individual statements ordered A, B, C, D from top to the bottom.
– PERTURBATIONFLOW
Aug 3 at 13:23
math formatting tutorial/reference
– Omnomnomnom
Aug 3 at 13:26
please take the time to format properly the question using either mathjax or latex
– Ahmad Bazzi
Aug 3 at 13:18
please take the time to format properly the question using either mathjax or latex
– Ahmad Bazzi
Aug 3 at 13:18
@AhmadBazzi is there a tutorial on using mathjax or latex?
– PERTURBATIONFLOW
Aug 3 at 13:18
@AhmadBazzi is there a tutorial on using mathjax or latex?
– PERTURBATIONFLOW
Aug 3 at 13:18
and also is it statement or statement(s) ?
– Ahmad Bazzi
Aug 3 at 13:21
and also is it statement or statement(s) ?
– Ahmad Bazzi
Aug 3 at 13:21
@AhmadBazzi it's a group of individual statements ordered A, B, C, D from top to the bottom.
– PERTURBATIONFLOW
Aug 3 at 13:23
@AhmadBazzi it's a group of individual statements ordered A, B, C, D from top to the bottom.
– PERTURBATIONFLOW
Aug 3 at 13:23
math formatting tutorial/reference
– Omnomnomnom
Aug 3 at 13:26
math formatting tutorial/reference
– Omnomnomnom
Aug 3 at 13:26
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
You're right about your assertions. Let me just give some notes on them as well:
For (b), you might think about the identity, certainly invertible but with every vector an eigenvector of $1$(besides $mathbf0$ as the null-vector never is an eigenvector).
For (c), as you rightly remarked, you have that for a matrix $Z$, $0$ is an eigenvalue of $Z$ iff $p_Z(0)=0$ iff $det(Z-0E_n)=0$ iff $det(Z)=0$ iff $Z$ is singular, i.e. not invertible.
For (d), the rank-nullity theorem seems to be the best way to go.
Now for (a), note that for $Z=(z_1,dots,z_n)$ being the column representation of $Z$, we have that if $z_1,dots,z_n$ span $mathbbR^n$, then f.a $vinmathbbR^n$, there exists $x_1,dots,x_ninmathbbR$ s.t. $sum_i=1^nx_iz_i=v$. Thus, using matrix multiplication, for every $vinmathbbR^n$, there exists an $xinmathbbR^n$ s.t. $Zx=v$, as $Zx=sum_i=1^nx_iz_i$ for $x=(x_1,dots,x_n)$.
Thus, the linear map $phi_Z(x)=Zx$ for $xinmathbbR^n$ is surjective. Now, by the rank-nullity theorem, a linear map between two spaces with the same dimension (as here $mathbbR^n$) is surjective iff it is injective iff it is bijective. Note that this only holds in finite dimensions. Thus $phi_Z$ is bijective and thus $Z$ is invertible.
To round this off, in your case of the finite dimensional $mathbbR^n$ you have $Z$ is invertible iff the columns of $Z$ span $mathbbR^n$ iff the columns of $Z$ are linearly independent. The last of which would correspond by a similar argument as above to injectivity(check this yourself) and thus we'd apply the argument again that this implies bijectivity rightout.
answered Aug 3 at 13:29
zzuussee
1,101419
add a comment |Â
up vote
1
down vote
A is true. $mathbb R^n$ is an $n$-dimensional vector space, so if $n$ vectors span $mathbb R^n$ then these $n$ vectors are a basis of $mathbb R^n$. Hence the columns of the matrix $Z$ span $mathbb R^n$ iff the columns are linearly independent, iff the system of equations $Zx=0$ has only the trivial solution, iff $Z$ is invertible.
answered Aug 3 at 13:27
Mark
5949
add a comment |Â
up vote
1
down vote
We have
- That the columns of Z span R^n
Yes indeed in that case the matrix is full rank and invertible.
- That Z has no eigenvectors
This point does not make sense.
- That Z has no eigenvectors of eigenvalue = 0
In that case A is not invertible since $det(A)=prod(lambda_i)=0$
- Kernel of Z = 0
Yes indeed that implies that columns are independent ($A$ is full rank).
answered Aug 3 at 13:31
gimusi
63.7k73480
add a comment |Â
up vote
0
down vote
The statements follow the Invertible Matrix Theorem.. The 1st, 3rd and 4th all work but the 2nd doesn't make much sense.
A is true because if the columns span $mathbbR^n$ they are linearly independent then invertible
answered Aug 3 at 13:26
RHowe
803715
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You're right about your assertions. Let me just give some notes on them as well:
For (b), you might think about the identity, certainly invertible but with every vector an eigenvector of $1$(besides $mathbf0$ as the null-vector never is an eigenvector).
For (c), as you rightly remarked, you have that for a matrix $Z$, $0$ is an eigenvalue of $Z$ iff $p_Z(0)=0$ iff $det(Z-0E_n)=0$ iff $det(Z)=0$ iff $Z$ is singular, i.e. not invertible.
For (d), the rank-nullity theorem seems to be the best way to go.
Now for (a), note that for $Z=(z_1,dots,z_n)$ being the column representation of $Z$, we have that if $z_1,dots,z_n$ span $mathbbR^n$, then f.a $vinmathbbR^n$, there exists $x_1,dots,x_ninmathbbR$ s.t. $sum_i=1^nx_iz_i=v$. Thus, using matrix multiplication, for every $vinmathbbR^n$, there exists an $xinmathbbR^n$ s.t. $Zx=v$, as $Zx=sum_i=1^nx_iz_i$ for $x=(x_1,dots,x_n)$.
Thus, the linear map $phi_Z(x)=Zx$ for $xinmathbbR^n$ is surjective. Now, by the rank-nullity theorem, a linear map between two spaces with the same dimension (as here $mathbbR^n$) is surjective iff it is injective iff it is bijective. Note that this only holds in finite dimensions. Thus $phi_Z$ is bijective and thus $Z$ is invertible.
To round this off, in your case of the finite dimensional $mathbbR^n$ you have $Z$ is invertible iff the columns of $Z$ span $mathbbR^n$ iff the columns of $Z$ are linearly independent. The last of which would correspond by a similar argument as above to injectivity(check this yourself) and thus we'd apply the argument again that this implies bijectivity rightout.
answered Aug 3 at 13:29
zzuussee
1,101419
add a comment |Â
up vote
1
down vote
accepted
You're right about your assertions. Let me just give some notes on them as well:
For (b), you might think about the identity, certainly invertible but with every vector an eigenvector of $1$(besides $mathbf0$ as the null-vector never is an eigenvector).
For (c), as you rightly remarked, you have that for a matrix $Z$, $0$ is an eigenvalue of $Z$ iff $p_Z(0)=0$ iff $det(Z-0E_n)=0$ iff $det(Z)=0$ iff $Z$ is singular, i.e. not invertible.
For (d), the rank-nullity theorem seems to be the best way to go.
Now for (a), note that for $Z=(z_1,dots,z_n)$ being the column representation of $Z$, we have that if $z_1,dots,z_n$ span $mathbbR^n$, then f.a $vinmathbbR^n$, there exists $x_1,dots,x_ninmathbbR$ s.t. $sum_i=1^nx_iz_i=v$. Thus, using matrix multiplication, for every $vinmathbbR^n$, there exists an $xinmathbbR^n$ s.t. $Zx=v$, as $Zx=sum_i=1^nx_iz_i$ for $x=(x_1,dots,x_n)$.
Thus, the linear map $phi_Z(x)=Zx$ for $xinmathbbR^n$ is surjective. Now, by the rank-nullity theorem, a linear map between two spaces with the same dimension (as here $mathbbR^n$) is surjective iff it is injective iff it is bijective. Note that this only holds in finite dimensions. Thus $phi_Z$ is bijective and thus $Z$ is invertible.
To round this off, in your case of the finite dimensional $mathbbR^n$ you have $Z$ is invertible iff the columns of $Z$ span $mathbbR^n$ iff the columns of $Z$ are linearly independent. The last of which would correspond by a similar argument as above to injectivity(check this yourself) and thus we'd apply the argument again that this implies bijectivity rightout.
answered Aug 3 at 13:29
zzuussee
1,101419
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You're right about your assertions. Let me just give some notes on them as well:
For (b), you might think about the identity, certainly invertible but with every vector an eigenvector of $1$(besides $mathbf0$ as the null-vector never is an eigenvector).
For (c), as you rightly remarked, you have that for a matrix $Z$, $0$ is an eigenvalue of $Z$ iff $p_Z(0)=0$ iff $det(Z-0E_n)=0$ iff $det(Z)=0$ iff $Z$ is singular, i.e. not invertible.
For (d), the rank-nullity theorem seems to be the best way to go.
Now for (a), note that for $Z=(z_1,dots,z_n)$ being the column representation of $Z$, we have that if $z_1,dots,z_n$ span $mathbbR^n$, then f.a $vinmathbbR^n$, there exists $x_1,dots,x_ninmathbbR$ s.t. $sum_i=1^nx_iz_i=v$. Thus, using matrix multiplication, for every $vinmathbbR^n$, there exists an $xinmathbbR^n$ s.t. $Zx=v$, as $Zx=sum_i=1^nx_iz_i$ for $x=(x_1,dots,x_n)$.
Thus, the linear map $phi_Z(x)=Zx$ for $xinmathbbR^n$ is surjective. Now, by the rank-nullity theorem, a linear map between two spaces with the same dimension (as here $mathbbR^n$) is surjective iff it is injective iff it is bijective. Note that this only holds in finite dimensions. Thus $phi_Z$ is bijective and thus $Z$ is invertible.
To round this off, in your case of the finite dimensional $mathbbR^n$ you have $Z$ is invertible iff the columns of $Z$ span $mathbbR^n$ iff the columns of $Z$ are linearly independent. The last of which would correspond by a similar argument as above to injectivity(check this yourself) and thus we'd apply the argument again that this implies bijectivity rightout.
answered Aug 3 at 13:29
zzuussee
1,101419
You're right about your assertions. Let me just give some notes on them as well:
For (b), you might think about the identity, certainly invertible but with every vector an eigenvector of $1$(besides $mathbf0$ as the null-vector never is an eigenvector).
For (c), as you rightly remarked, you have that for a matrix $Z$, $0$ is an eigenvalue of $Z$ iff $p_Z(0)=0$ iff $det(Z-0E_n)=0$ iff $det(Z)=0$ iff $Z$ is singular, i.e. not invertible.
For (d), the rank-nullity theorem seems to be the best way to go.
Now for (a), note that for $Z=(z_1,dots,z_n)$ being the column representation of $Z$, we have that if $z_1,dots,z_n$ span $mathbbR^n$, then f.a $vinmathbbR^n$, there exists $x_1,dots,x_ninmathbbR$ s.t. $sum_i=1^nx_iz_i=v$. Thus, using matrix multiplication, for every $vinmathbbR^n$, there exists an $xinmathbbR^n$ s.t. $Zx=v$, as $Zx=sum_i=1^nx_iz_i$ for $x=(x_1,dots,x_n)$.
Thus, the linear map $phi_Z(x)=Zx$ for $xinmathbbR^n$ is surjective. Now, by the rank-nullity theorem, a linear map between two spaces with the same dimension (as here $mathbbR^n$) is surjective iff it is injective iff it is bijective. Note that this only holds in finite dimensions. Thus $phi_Z$ is bijective and thus $Z$ is invertible.
To round this off, in your case of the finite dimensional $mathbbR^n$ you have $Z$ is invertible iff the columns of $Z$ span $mathbbR^n$ iff the columns of $Z$ are linearly independent. The last of which would correspond by a similar argument as above to injectivity(check this yourself) and thus we'd apply the argument again that this implies bijectivity rightout.
answered Aug 3 at 13:29
zzuussee
1,101419
answered Aug 3 at 13:29
zzuussee
1,101419
answered Aug 3 at 13:29
zzuussee
1,101419
answered Aug 3 at 13:29
zzuussee
1,101419
1,101419
add a comment |Â
add a comment |Â
up vote
1
down vote
A is true. $mathbb R^n$ is an $n$-dimensional vector space, so if $n$ vectors span $mathbb R^n$ then these $n$ vectors are a basis of $mathbb R^n$. Hence the columns of the matrix $Z$ span $mathbb R^n$ iff the columns are linearly independent, iff the system of equations $Zx=0$ has only the trivial solution, iff $Z$ is invertible.
answered Aug 3 at 13:27
Mark
5949
add a comment |Â
up vote
1
down vote
A is true. $mathbb R^n$ is an $n$-dimensional vector space, so if $n$ vectors span $mathbb R^n$ then these $n$ vectors are a basis of $mathbb R^n$. Hence the columns of the matrix $Z$ span $mathbb R^n$ iff the columns are linearly independent, iff the system of equations $Zx=0$ has only the trivial solution, iff $Z$ is invertible.
answered Aug 3 at 13:27
Mark
5949
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A is true. $mathbb R^n$ is an $n$-dimensional vector space, so if $n$ vectors span $mathbb R^n$ then these $n$ vectors are a basis of $mathbb R^n$. Hence the columns of the matrix $Z$ span $mathbb R^n$ iff the columns are linearly independent, iff the system of equations $Zx=0$ has only the trivial solution, iff $Z$ is invertible.
answered Aug 3 at 13:27
Mark
5949
A is true. $mathbb R^n$ is an $n$-dimensional vector space, so if $n$ vectors span $mathbb R^n$ then these $n$ vectors are a basis of $mathbb R^n$. Hence the columns of the matrix $Z$ span $mathbb R^n$ iff the columns are linearly independent, iff the system of equations $Zx=0$ has only the trivial solution, iff $Z$ is invertible.
answered Aug 3 at 13:27
Mark
5949
answered Aug 3 at 13:27
Mark
5949
answered Aug 3 at 13:27
Mark
5949
answered Aug 3 at 13:27
Mark
5949
5949
add a comment |Â
add a comment |Â
up vote
1
down vote
We have
- That the columns of Z span R^n
Yes indeed in that case the matrix is full rank and invertible.
- That Z has no eigenvectors
This point does not make sense.
- That Z has no eigenvectors of eigenvalue = 0
In that case A is not invertible since $det(A)=prod(lambda_i)=0$
- Kernel of Z = 0
Yes indeed that implies that columns are independent ($A$ is full rank).
answered Aug 3 at 13:31
gimusi
63.7k73480
add a comment |Â
up vote
1
down vote
We have
- That the columns of Z span R^n
Yes indeed in that case the matrix is full rank and invertible.
- That Z has no eigenvectors
This point does not make sense.
- That Z has no eigenvectors of eigenvalue = 0
In that case A is not invertible since $det(A)=prod(lambda_i)=0$
- Kernel of Z = 0
Yes indeed that implies that columns are independent ($A$ is full rank).
answered Aug 3 at 13:31
gimusi
63.7k73480
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We have
- That the columns of Z span R^n
Yes indeed in that case the matrix is full rank and invertible.
- That Z has no eigenvectors
This point does not make sense.
- That Z has no eigenvectors of eigenvalue = 0
In that case A is not invertible since $det(A)=prod(lambda_i)=0$
- Kernel of Z = 0
Yes indeed that implies that columns are independent ($A$ is full rank).
answered Aug 3 at 13:31
gimusi
63.7k73480
We have
- That the columns of Z span R^n
Yes indeed in that case the matrix is full rank and invertible.
- That Z has no eigenvectors
This point does not make sense.
- That Z has no eigenvectors of eigenvalue = 0
In that case A is not invertible since $det(A)=prod(lambda_i)=0$
- Kernel of Z = 0
Yes indeed that implies that columns are independent ($A$ is full rank).
answered Aug 3 at 13:31
gimusi
63.7k73480
answered Aug 3 at 13:31
gimusi
63.7k73480
answered Aug 3 at 13:31
gimusi
63.7k73480
answered Aug 3 at 13:31
gimusi
63.7k73480
63.7k73480
add a comment |Â
add a comment |Â
up vote
0
down vote
The statements follow the Invertible Matrix Theorem.. The 1st, 3rd and 4th all work but the 2nd doesn't make much sense.
A is true because if the columns span $mathbbR^n$ they are linearly independent then invertible
answered Aug 3 at 13:26
RHowe
803715
add a comment |Â
up vote
0
down vote
The statements follow the Invertible Matrix Theorem.. The 1st, 3rd and 4th all work but the 2nd doesn't make much sense.
A is true because if the columns span $mathbbR^n$ they are linearly independent then invertible
answered Aug 3 at 13:26
RHowe
803715
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The statements follow the Invertible Matrix Theorem.. The 1st, 3rd and 4th all work but the 2nd doesn't make much sense.
A is true because if the columns span $mathbbR^n$ they are linearly independent then invertible
answered Aug 3 at 13:26
RHowe
803715
The statements follow the Invertible Matrix Theorem.. The 1st, 3rd and 4th all work but the 2nd doesn't make much sense.
A is true because if the columns span $mathbbR^n$ they are linearly independent then invertible
answered Aug 3 at 13:26
RHowe
803715
answered Aug 3 at 13:26
RHowe
803715
answered Aug 3 at 13:26
RHowe
803715
answered Aug 3 at 13:26
RHowe
803715
803715
add a comment |Â
add a comment |Â
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please take the time to format properly the question using either mathjax or latex
– Ahmad Bazzi
Aug 3 at 13:18
@AhmadBazzi is there a tutorial on using mathjax or latex?
– PERTURBATIONFLOW
Aug 3 at 13:18
and also is it statement or statement(s) ?
– Ahmad Bazzi
Aug 3 at 13:21
@AhmadBazzi it's a group of individual statements ordered A, B, C, D from top to the bottom.
– PERTURBATIONFLOW
Aug 3 at 13:23
math formatting tutorial/reference
– Omnomnomnom
Aug 3 at 13:26