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Show that $T$ is a linear transformation and find a basis for $N(T)$
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Consider the vector space of 2-by-2 matrices $M_2times 2(mathbbR)$ over $mathbbR$. For a given matrix A=$beginpmatrixa_1,1 &a_1,2\a_2,1&a_2,2endpmatrix$, consider the map $T:M_2times 2(mathbbR) to mathbbR$ given by $T(A)=a_2,1$
$a)$ Show that T is a linear transformation.
Definition: Let $V$ and $W$ be vector spaces over $F$. We call a function $T : V to W$ a linear
transformation from $V$ to $W$ if, for all $x, y in V$ and $c in F$, we have
$(1) T (x + y) = T (x) + T (y)$
$(2) T (cx) = cT (x)$
Using this definition this is the solution I came up with
$T(cbeginpmatrixa_1,1 &a_1,2\a_2,1&a_2,2endpmatrix+beginpmatrixb_1,1 &b_1,2\b_2,1&b_2,2endpmatrix) =c(a_2,1)+(b_2,1)=cT(A)+T(B)$
$b)$ Find a basis for $N(T)$
Definition: Null Space $N(T )=x in V : T (x) = 0$.
Not sure where to begin with this. The following questions I assume can only be solved if finding a solution to $b)$
$c)$ Calculate the dimension of $N(T)$.
$d)$ Find a basis for $R(T)$
Thanks in advance.
linear-algebra vector-spaces
edited Jul 25 at 11:58
Kenta S
1,1371418
asked Jul 25 at 10:54
Ben Jones
17911
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Consider the vector space of 2-by-2 matrices $M_2times 2(mathbbR)$ over $mathbbR$. For a given matrix A=$beginpmatrixa_1,1 &a_1,2\a_2,1&a_2,2endpmatrix$, consider the map $T:M_2times 2(mathbbR) to mathbbR$ given by $T(A)=a_2,1$
$a)$ Show that T is a linear transformation.
Definition: Let $V$ and $W$ be vector spaces over $F$. We call a function $T : V to W$ a linear
transformation from $V$ to $W$ if, for all $x, y in V$ and $c in F$, we have
$(1) T (x + y) = T (x) + T (y)$
$(2) T (cx) = cT (x)$
Using this definition this is the solution I came up with
$T(cbeginpmatrixa_1,1 &a_1,2\a_2,1&a_2,2endpmatrix+beginpmatrixb_1,1 &b_1,2\b_2,1&b_2,2endpmatrix) =c(a_2,1)+(b_2,1)=cT(A)+T(B)$
$b)$ Find a basis for $N(T)$
Definition: Null Space $N(T )=x in V : T (x) = 0$.
Not sure where to begin with this. The following questions I assume can only be solved if finding a solution to $b)$
$c)$ Calculate the dimension of $N(T)$.
$d)$ Find a basis for $R(T)$
Thanks in advance.
linear-algebra vector-spaces
edited Jul 25 at 11:58
Kenta S
1,1371418
asked Jul 25 at 10:54
Ben Jones
17911
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the vector space of 2-by-2 matrices $M_2times 2(mathbbR)$ over $mathbbR$. For a given matrix A=$beginpmatrixa_1,1 &a_1,2\a_2,1&a_2,2endpmatrix$, consider the map $T:M_2times 2(mathbbR) to mathbbR$ given by $T(A)=a_2,1$
$a)$ Show that T is a linear transformation.
Definition: Let $V$ and $W$ be vector spaces over $F$. We call a function $T : V to W$ a linear
transformation from $V$ to $W$ if, for all $x, y in V$ and $c in F$, we have
$(1) T (x + y) = T (x) + T (y)$
$(2) T (cx) = cT (x)$
Using this definition this is the solution I came up with
$T(cbeginpmatrixa_1,1 &a_1,2\a_2,1&a_2,2endpmatrix+beginpmatrixb_1,1 &b_1,2\b_2,1&b_2,2endpmatrix) =c(a_2,1)+(b_2,1)=cT(A)+T(B)$
$b)$ Find a basis for $N(T)$
Definition: Null Space $N(T )=x in V : T (x) = 0$.
Not sure where to begin with this. The following questions I assume can only be solved if finding a solution to $b)$
$c)$ Calculate the dimension of $N(T)$.
$d)$ Find a basis for $R(T)$
Thanks in advance.
linear-algebra vector-spaces
edited Jul 25 at 11:58
Kenta S
1,1371418
asked Jul 25 at 10:54
Ben Jones
17911
Consider the vector space of 2-by-2 matrices $M_2times 2(mathbbR)$ over $mathbbR$. For a given matrix A=$beginpmatrixa_1,1 &a_1,2\a_2,1&a_2,2endpmatrix$, consider the map $T:M_2times 2(mathbbR) to mathbbR$ given by $T(A)=a_2,1$
$a)$ Show that T is a linear transformation.
Definition: Let $V$ and $W$ be vector spaces over $F$. We call a function $T : V to W$ a linear
transformation from $V$ to $W$ if, for all $x, y in V$ and $c in F$, we have
$(1) T (x + y) = T (x) + T (y)$
$(2) T (cx) = cT (x)$
Using this definition this is the solution I came up with
$T(cbeginpmatrixa_1,1 &a_1,2\a_2,1&a_2,2endpmatrix+beginpmatrixb_1,1 &b_1,2\b_2,1&b_2,2endpmatrix) =c(a_2,1)+(b_2,1)=cT(A)+T(B)$
$b)$ Find a basis for $N(T)$
Definition: Null Space $N(T )=x in V : T (x) = 0$.
Not sure where to begin with this. The following questions I assume can only be solved if finding a solution to $b)$
$c)$ Calculate the dimension of $N(T)$.
$d)$ Find a basis for $R(T)$
Thanks in advance.
linear-algebra vector-spaces
edited Jul 25 at 11:58
Kenta S
1,1371418
asked Jul 25 at 10:54
Ben Jones
17911
edited Jul 25 at 11:58
Kenta S
1,1371418
edited Jul 25 at 11:58
Kenta S
1,1371418
edited Jul 25 at 11:58
Kenta S
1,1371418
1,1371418
asked Jul 25 at 10:54
Ben Jones
17911
asked Jul 25 at 10:54
Ben Jones
17911
asked Jul 25 at 10:54
Ben Jones
17911
17911
add a comment |Â
add a comment |Â
3 Answers
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active
oldest
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up vote
1
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accepted
You solution to part (a) seems to be just fine. For part (b):
Consider what is needed for a matrix $AinmathbbR^(2,2)$ to map to $0$ under $T$, i.e. $T(A)=0$, i.e. $a_2,1=0$. Thus
$$
N(T)=AinmathbbR^(2,2)mid a_2,1=0=beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrixinmathbbR^(2,2)
$$
The classical basis for $mathbbR^(2,2)$ is the basis associated with the canonical basis for $mathbbR^4$:
$$
beginpmatrix1 &0\0 &0endpmatrix, beginpmatrix0 &1\0 &0endpmatrix,beginpmatrix0 &0\1 &0endpmatrix, beginpmatrix0 &0\0 &1endpmatrix
$$
You may check that these are indeed linear independent and span $mathbbR^(2,2)$. Coming back to the null space of $T$, you can see that every matrix $Ain N(T)$, i.e. every matrix of the form
$$
beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix
$$
can be created as a linear combination of
$$
beginpmatrix1 &0\0 &0endpmatrix, beginpmatrix0 &1\0 &0endpmatrix,beginpmatrix0 &0\0 &1endpmatrix
$$
with
$$
beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix=a_1,1beginpmatrix1 &0\0 &0endpmatrix + a_1,2 beginpmatrix0 &1\0 &0endpmatrix+ a_2,2beginpmatrix0 &0\0 &1endpmatrix
$$
as the defining condition for $N(T)$ is a zero at position $a_2,1$. As we found a set of linear independent vectors(the three matrices above) that span $N(T)$, we've found a basis for the concerned space. Followingly, as an answer to part (c), we find that $mathrmdim(N(T))=3$.
Note, that by the rank-nullity theorem(and assuming that $R(T)$ means the range of $T$), we have that
$$
4=mathrmdim(mathbbR^(2,2))=mathrmdim(N(T))+mathrmdim(R(T))=3+mathrmdim(R(T))
$$
Therefore, $mathrmdim(R(T))=1$, i.e. as $T:mathbbR^(2,2)tomathbbR$, we have $R(T)=mathbbR$. A typical basis for $mathbbR$ as a vector space is the number $1$.
answered Jul 25 at 11:16
zzuussee
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Let $A_1= beginpmatrix1 &0\0 &0endpmatrix$, $A_2= beginpmatrix0 &1\0 &0endpmatrix$ and $A_3= beginpmatrix0 &0\0 &1endpmatrix$, then
$A in N(T) iff A=beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix iff $
$A in span A_1,A_2,A_3 $.
Can you proceed ?
answered Jul 25 at 11:04
Fred
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a. Let $X=[x_i,j],Y=[y_i,j]$ be 2x2 matrices and $cin F$.
(1) $T(X+Y) = x_2,1 + y_2,1 = T(X)+T(Y)$
(2)$T(cX) = cx_2,1 = cT(X). $
This proves that T is a linear trasformation.
b. $N(T) = leftlbrace A = [a_i,j] | a_2,1=0 rightrbrace
= Spanleft( leftlbrace left[ beginarraycc 1&0\0&0endarrayright] , left[ beginarraycc 0&1\0&0endarrayright],left[ beginarraycc 0&0\0&1endarrayright] rightrbrace right)$
Show that $leftlbrace left[ beginarraycc 1&0\0&0endarrayright] , left[ beginarraycc 0&1\0&0endarrayright],left[ beginarraycc 0&0\0&1endarrayright] rightrbrace$ is linearly independent to how that it is a basis.
answered Jul 25 at 11:13
Jared S.
1
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You solution to part (a) seems to be just fine. For part (b):
Consider what is needed for a matrix $AinmathbbR^(2,2)$ to map to $0$ under $T$, i.e. $T(A)=0$, i.e. $a_2,1=0$. Thus
$$
N(T)=AinmathbbR^(2,2)mid a_2,1=0=beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrixinmathbbR^(2,2)
$$
The classical basis for $mathbbR^(2,2)$ is the basis associated with the canonical basis for $mathbbR^4$:
$$
beginpmatrix1 &0\0 &0endpmatrix, beginpmatrix0 &1\0 &0endpmatrix,beginpmatrix0 &0\1 &0endpmatrix, beginpmatrix0 &0\0 &1endpmatrix
$$
You may check that these are indeed linear independent and span $mathbbR^(2,2)$. Coming back to the null space of $T$, you can see that every matrix $Ain N(T)$, i.e. every matrix of the form
$$
beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix
$$
can be created as a linear combination of
$$
beginpmatrix1 &0\0 &0endpmatrix, beginpmatrix0 &1\0 &0endpmatrix,beginpmatrix0 &0\0 &1endpmatrix
$$
with
$$
beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix=a_1,1beginpmatrix1 &0\0 &0endpmatrix + a_1,2 beginpmatrix0 &1\0 &0endpmatrix+ a_2,2beginpmatrix0 &0\0 &1endpmatrix
$$
as the defining condition for $N(T)$ is a zero at position $a_2,1$. As we found a set of linear independent vectors(the three matrices above) that span $N(T)$, we've found a basis for the concerned space. Followingly, as an answer to part (c), we find that $mathrmdim(N(T))=3$.
Note, that by the rank-nullity theorem(and assuming that $R(T)$ means the range of $T$), we have that
$$
4=mathrmdim(mathbbR^(2,2))=mathrmdim(N(T))+mathrmdim(R(T))=3+mathrmdim(R(T))
$$
Therefore, $mathrmdim(R(T))=1$, i.e. as $T:mathbbR^(2,2)tomathbbR$, we have $R(T)=mathbbR$. A typical basis for $mathbbR$ as a vector space is the number $1$.
answered Jul 25 at 11:16
zzuussee
1,514419
add a comment |Â
up vote
1
down vote
accepted
You solution to part (a) seems to be just fine. For part (b):
Consider what is needed for a matrix $AinmathbbR^(2,2)$ to map to $0$ under $T$, i.e. $T(A)=0$, i.e. $a_2,1=0$. Thus
$$
N(T)=AinmathbbR^(2,2)mid a_2,1=0=beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrixinmathbbR^(2,2)
$$
The classical basis for $mathbbR^(2,2)$ is the basis associated with the canonical basis for $mathbbR^4$:
$$
beginpmatrix1 &0\0 &0endpmatrix, beginpmatrix0 &1\0 &0endpmatrix,beginpmatrix0 &0\1 &0endpmatrix, beginpmatrix0 &0\0 &1endpmatrix
$$
You may check that these are indeed linear independent and span $mathbbR^(2,2)$. Coming back to the null space of $T$, you can see that every matrix $Ain N(T)$, i.e. every matrix of the form
$$
beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix
$$
can be created as a linear combination of
$$
beginpmatrix1 &0\0 &0endpmatrix, beginpmatrix0 &1\0 &0endpmatrix,beginpmatrix0 &0\0 &1endpmatrix
$$
with
$$
beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix=a_1,1beginpmatrix1 &0\0 &0endpmatrix + a_1,2 beginpmatrix0 &1\0 &0endpmatrix+ a_2,2beginpmatrix0 &0\0 &1endpmatrix
$$
as the defining condition for $N(T)$ is a zero at position $a_2,1$. As we found a set of linear independent vectors(the three matrices above) that span $N(T)$, we've found a basis for the concerned space. Followingly, as an answer to part (c), we find that $mathrmdim(N(T))=3$.
Note, that by the rank-nullity theorem(and assuming that $R(T)$ means the range of $T$), we have that
$$
4=mathrmdim(mathbbR^(2,2))=mathrmdim(N(T))+mathrmdim(R(T))=3+mathrmdim(R(T))
$$
Therefore, $mathrmdim(R(T))=1$, i.e. as $T:mathbbR^(2,2)tomathbbR$, we have $R(T)=mathbbR$. A typical basis for $mathbbR$ as a vector space is the number $1$.
answered Jul 25 at 11:16
zzuussee
1,514419
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You solution to part (a) seems to be just fine. For part (b):
Consider what is needed for a matrix $AinmathbbR^(2,2)$ to map to $0$ under $T$, i.e. $T(A)=0$, i.e. $a_2,1=0$. Thus
$$
N(T)=AinmathbbR^(2,2)mid a_2,1=0=beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrixinmathbbR^(2,2)
$$
The classical basis for $mathbbR^(2,2)$ is the basis associated with the canonical basis for $mathbbR^4$:
$$
beginpmatrix1 &0\0 &0endpmatrix, beginpmatrix0 &1\0 &0endpmatrix,beginpmatrix0 &0\1 &0endpmatrix, beginpmatrix0 &0\0 &1endpmatrix
$$
You may check that these are indeed linear independent and span $mathbbR^(2,2)$. Coming back to the null space of $T$, you can see that every matrix $Ain N(T)$, i.e. every matrix of the form
$$
beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix
$$
can be created as a linear combination of
$$
beginpmatrix1 &0\0 &0endpmatrix, beginpmatrix0 &1\0 &0endpmatrix,beginpmatrix0 &0\0 &1endpmatrix
$$
with
$$
beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix=a_1,1beginpmatrix1 &0\0 &0endpmatrix + a_1,2 beginpmatrix0 &1\0 &0endpmatrix+ a_2,2beginpmatrix0 &0\0 &1endpmatrix
$$
as the defining condition for $N(T)$ is a zero at position $a_2,1$. As we found a set of linear independent vectors(the three matrices above) that span $N(T)$, we've found a basis for the concerned space. Followingly, as an answer to part (c), we find that $mathrmdim(N(T))=3$.
Note, that by the rank-nullity theorem(and assuming that $R(T)$ means the range of $T$), we have that
$$
4=mathrmdim(mathbbR^(2,2))=mathrmdim(N(T))+mathrmdim(R(T))=3+mathrmdim(R(T))
$$
Therefore, $mathrmdim(R(T))=1$, i.e. as $T:mathbbR^(2,2)tomathbbR$, we have $R(T)=mathbbR$. A typical basis for $mathbbR$ as a vector space is the number $1$.
answered Jul 25 at 11:16
zzuussee
1,514419
You solution to part (a) seems to be just fine. For part (b):
Consider what is needed for a matrix $AinmathbbR^(2,2)$ to map to $0$ under $T$, i.e. $T(A)=0$, i.e. $a_2,1=0$. Thus
$$
N(T)=AinmathbbR^(2,2)mid a_2,1=0=beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrixinmathbbR^(2,2)
$$
The classical basis for $mathbbR^(2,2)$ is the basis associated with the canonical basis for $mathbbR^4$:
$$
beginpmatrix1 &0\0 &0endpmatrix, beginpmatrix0 &1\0 &0endpmatrix,beginpmatrix0 &0\1 &0endpmatrix, beginpmatrix0 &0\0 &1endpmatrix
$$
You may check that these are indeed linear independent and span $mathbbR^(2,2)$. Coming back to the null space of $T$, you can see that every matrix $Ain N(T)$, i.e. every matrix of the form
$$
beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix
$$
can be created as a linear combination of
$$
beginpmatrix1 &0\0 &0endpmatrix, beginpmatrix0 &1\0 &0endpmatrix,beginpmatrix0 &0\0 &1endpmatrix
$$
with
$$
beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix=a_1,1beginpmatrix1 &0\0 &0endpmatrix + a_1,2 beginpmatrix0 &1\0 &0endpmatrix+ a_2,2beginpmatrix0 &0\0 &1endpmatrix
$$
as the defining condition for $N(T)$ is a zero at position $a_2,1$. As we found a set of linear independent vectors(the three matrices above) that span $N(T)$, we've found a basis for the concerned space. Followingly, as an answer to part (c), we find that $mathrmdim(N(T))=3$.
Note, that by the rank-nullity theorem(and assuming that $R(T)$ means the range of $T$), we have that
$$
4=mathrmdim(mathbbR^(2,2))=mathrmdim(N(T))+mathrmdim(R(T))=3+mathrmdim(R(T))
$$
Therefore, $mathrmdim(R(T))=1$, i.e. as $T:mathbbR^(2,2)tomathbbR$, we have $R(T)=mathbbR$. A typical basis for $mathbbR$ as a vector space is the number $1$.
answered Jul 25 at 11:16
zzuussee
1,514419
answered Jul 25 at 11:16
zzuussee
1,514419
answered Jul 25 at 11:16
zzuussee
1,514419
answered Jul 25 at 11:16
zzuussee
1,514419
1,514419
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $A_1= beginpmatrix1 &0\0 &0endpmatrix$, $A_2= beginpmatrix0 &1\0 &0endpmatrix$ and $A_3= beginpmatrix0 &0\0 &1endpmatrix$, then
$A in N(T) iff A=beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix iff $
$A in span A_1,A_2,A_3 $.
Can you proceed ?
answered Jul 25 at 11:04
Fred
37.2k1237
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up vote
0
down vote
Let $A_1= beginpmatrix1 &0\0 &0endpmatrix$, $A_2= beginpmatrix0 &1\0 &0endpmatrix$ and $A_3= beginpmatrix0 &0\0 &1endpmatrix$, then
$A in N(T) iff A=beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix iff $
$A in span A_1,A_2,A_3 $.
Can you proceed ?
answered Jul 25 at 11:04
Fred
37.2k1237
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $A_1= beginpmatrix1 &0\0 &0endpmatrix$, $A_2= beginpmatrix0 &1\0 &0endpmatrix$ and $A_3= beginpmatrix0 &0\0 &1endpmatrix$, then
$A in N(T) iff A=beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix iff $
$A in span A_1,A_2,A_3 $.
Can you proceed ?
answered Jul 25 at 11:04
Fred
37.2k1237
Let $A_1= beginpmatrix1 &0\0 &0endpmatrix$, $A_2= beginpmatrix0 &1\0 &0endpmatrix$ and $A_3= beginpmatrix0 &0\0 &1endpmatrix$, then
$A in N(T) iff A=beginpmatrixa_1,1 &a_1,2\0 &a_2,2endpmatrix iff $
$A in span A_1,A_2,A_3 $.
Can you proceed ?
answered Jul 25 at 11:04
Fred
37.2k1237
answered Jul 25 at 11:04
Fred
37.2k1237
answered Jul 25 at 11:04
Fred
37.2k1237
answered Jul 25 at 11:04
Fred
37.2k1237
37.2k1237
add a comment |Â
add a comment |Â
up vote
0
down vote
a. Let $X=[x_i,j],Y=[y_i,j]$ be 2x2 matrices and $cin F$.
(1) $T(X+Y) = x_2,1 + y_2,1 = T(X)+T(Y)$
(2)$T(cX) = cx_2,1 = cT(X). $
This proves that T is a linear trasformation.
b. $N(T) = leftlbrace A = [a_i,j] | a_2,1=0 rightrbrace
= Spanleft( leftlbrace left[ beginarraycc 1&0\0&0endarrayright] , left[ beginarraycc 0&1\0&0endarrayright],left[ beginarraycc 0&0\0&1endarrayright] rightrbrace right)$
Show that $leftlbrace left[ beginarraycc 1&0\0&0endarrayright] , left[ beginarraycc 0&1\0&0endarrayright],left[ beginarraycc 0&0\0&1endarrayright] rightrbrace$ is linearly independent to how that it is a basis.
answered Jul 25 at 11:13
Jared S.
1
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up vote
0
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a. Let $X=[x_i,j],Y=[y_i,j]$ be 2x2 matrices and $cin F$.
(1) $T(X+Y) = x_2,1 + y_2,1 = T(X)+T(Y)$
(2)$T(cX) = cx_2,1 = cT(X). $
This proves that T is a linear trasformation.
b. $N(T) = leftlbrace A = [a_i,j] | a_2,1=0 rightrbrace
= Spanleft( leftlbrace left[ beginarraycc 1&0\0&0endarrayright] , left[ beginarraycc 0&1\0&0endarrayright],left[ beginarraycc 0&0\0&1endarrayright] rightrbrace right)$
Show that $leftlbrace left[ beginarraycc 1&0\0&0endarrayright] , left[ beginarraycc 0&1\0&0endarrayright],left[ beginarraycc 0&0\0&1endarrayright] rightrbrace$ is linearly independent to how that it is a basis.
answered Jul 25 at 11:13
Jared S.
1
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up vote
0
down vote
up vote
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down vote
a. Let $X=[x_i,j],Y=[y_i,j]$ be 2x2 matrices and $cin F$.
(1) $T(X+Y) = x_2,1 + y_2,1 = T(X)+T(Y)$
(2)$T(cX) = cx_2,1 = cT(X). $
This proves that T is a linear trasformation.
b. $N(T) = leftlbrace A = [a_i,j] | a_2,1=0 rightrbrace
= Spanleft( leftlbrace left[ beginarraycc 1&0\0&0endarrayright] , left[ beginarraycc 0&1\0&0endarrayright],left[ beginarraycc 0&0\0&1endarrayright] rightrbrace right)$
Show that $leftlbrace left[ beginarraycc 1&0\0&0endarrayright] , left[ beginarraycc 0&1\0&0endarrayright],left[ beginarraycc 0&0\0&1endarrayright] rightrbrace$ is linearly independent to how that it is a basis.
answered Jul 25 at 11:13
Jared S.
1
a. Let $X=[x_i,j],Y=[y_i,j]$ be 2x2 matrices and $cin F$.
(1) $T(X+Y) = x_2,1 + y_2,1 = T(X)+T(Y)$
(2)$T(cX) = cx_2,1 = cT(X). $
This proves that T is a linear trasformation.
b. $N(T) = leftlbrace A = [a_i,j] | a_2,1=0 rightrbrace
= Spanleft( leftlbrace left[ beginarraycc 1&0\0&0endarrayright] , left[ beginarraycc 0&1\0&0endarrayright],left[ beginarraycc 0&0\0&1endarrayright] rightrbrace right)$
Show that $leftlbrace left[ beginarraycc 1&0\0&0endarrayright] , left[ beginarraycc 0&1\0&0endarrayright],left[ beginarraycc 0&0\0&1endarrayright] rightrbrace$ is linearly independent to how that it is a basis.
answered Jul 25 at 11:13
Jared S.
1
answered Jul 25 at 11:13
Jared S.
1
answered Jul 25 at 11:13
Jared S.
1
answered Jul 25 at 11:13
Jared S.
1
1
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