Find the associated matrix of a linear transformationEigenvalues and eigenvectors of a matrix-transformationFind the standard matrix representation of the linear transformation T in M2,2Find the matrix associated to a linear transformationFind matrix of linear transformation $mathcalA$Find matrix A of the linear transformationFind the representative matrix for a linear transformationLinear transformation in matrix spaceMatrix associated to a linear transformation.What is the intuitive meaning of left multiply elemantary matrix as a linear transformation?Matrix associated of a Linear TransformationMatrix Representation of Linear Transformation from R2x2 to R3
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Find the associated matrix of a linear transformation
Eigenvalues and eigenvectors of a matrix-transformationFind the standard matrix representation of the linear transformation T in M2,2Find the matrix associated to a linear transformationFind matrix of linear transformation $mathcalA$Find matrix A of the linear transformationFind the representative matrix for a linear transformationLinear transformation in matrix spaceMatrix associated to a linear transformation.What is the intuitive meaning of left multiply elemantary matrix as a linear transformation?Matrix associated of a Linear TransformationMatrix Representation of Linear Transformation from R2x2 to R3
$begingroup$
Suppose
$$mathrmTleft(beginbmatrix
a & b \
c & d
endbmatrixright) = beginbmatrix
d & -b \
-c & a
endbmatrix,$$ can we determine the corresponding matrix of this linear transformation? Is it
$$
beginbmatrix
-1 & 0 \
0 & -1
endbmatrix?
$$
linear-algebra
edited Mar 21 at 18:55
Daniele Tampieri
2,65221022
asked Mar 21 at 18:36
EricEric
386
$endgroup$
add a comment |
$begingroup$
Suppose
$$mathrmTleft(beginbmatrix
a & b \
c & d
endbmatrixright) = beginbmatrix
d & -b \
-c & a
endbmatrix,$$ can we determine the corresponding matrix of this linear transformation? Is it
$$
beginbmatrix
-1 & 0 \
0 & -1
endbmatrix?
$$
linear-algebra
edited Mar 21 at 18:55
Daniele Tampieri
2,65221022
asked Mar 21 at 18:36
EricEric
386
$endgroup$
1
$begingroup$
With respect to which basis?
$endgroup$
– José Carlos Santos
Mar 21 at 18:37
$begingroup$
I just did one of these a minute ago math.stackexchange.com/questions/3157130/…
$endgroup$
– Will Jagy
Mar 21 at 18:40
1
$begingroup$
The space of $2times2$ matrix has dimension $4$, so the matrix of $T$ must be $4times4$.
$endgroup$
– TheSilverDoe
Mar 21 at 18:41
$begingroup$
but how could we multiply a 4*4 matrix with a 2*2 matrix?
$endgroup$
– Eric
Mar 21 at 19:39
add a comment |
$begingroup$
Suppose
$$mathrmTleft(beginbmatrix
a & b \
c & d
endbmatrixright) = beginbmatrix
d & -b \
-c & a
endbmatrix,$$ can we determine the corresponding matrix of this linear transformation? Is it
$$
beginbmatrix
-1 & 0 \
0 & -1
endbmatrix?
$$
linear-algebra
edited Mar 21 at 18:55
Daniele Tampieri
2,65221022
asked Mar 21 at 18:36
EricEric
386
$endgroup$
Suppose
$$mathrmTleft(beginbmatrix
a & b \
c & d
endbmatrixright) = beginbmatrix
d & -b \
-c & a
endbmatrix,$$ can we determine the corresponding matrix of this linear transformation? Is it
$$
beginbmatrix
-1 & 0 \
0 & -1
endbmatrix?
$$
linear-algebra
linear-algebra
edited Mar 21 at 18:55
Daniele Tampieri
2,65221022
asked Mar 21 at 18:36
EricEric
386
edited Mar 21 at 18:55
Daniele Tampieri
2,65221022
asked Mar 21 at 18:36
EricEric
386
edited Mar 21 at 18:55
Daniele Tampieri
2,65221022
edited Mar 21 at 18:55
Daniele Tampieri
2,65221022
edited Mar 21 at 18:55
Daniele Tampieri
2,65221022
2,65221022
asked Mar 21 at 18:36
EricEric
386
asked Mar 21 at 18:36
EricEric
386
asked Mar 21 at 18:36
EricEric
386
386
1
$begingroup$
With respect to which basis?
$endgroup$
– José Carlos Santos
Mar 21 at 18:37
$begingroup$
I just did one of these a minute ago math.stackexchange.com/questions/3157130/…
$endgroup$
– Will Jagy
Mar 21 at 18:40
1
$begingroup$
The space of $2times2$ matrix has dimension $4$, so the matrix of $T$ must be $4times4$.
$endgroup$
– TheSilverDoe
Mar 21 at 18:41
$begingroup$
but how could we multiply a 4*4 matrix with a 2*2 matrix?
$endgroup$
– Eric
Mar 21 at 19:39
add a comment |
1
$begingroup$
With respect to which basis?
$endgroup$
– José Carlos Santos
Mar 21 at 18:37
$begingroup$
I just did one of these a minute ago math.stackexchange.com/questions/3157130/…
$endgroup$
– Will Jagy
Mar 21 at 18:40
1
$begingroup$
The space of $2times2$ matrix has dimension $4$, so the matrix of $T$ must be $4times4$.
$endgroup$
– TheSilverDoe
Mar 21 at 18:41
$begingroup$
but how could we multiply a 4*4 matrix with a 2*2 matrix?
$endgroup$
– Eric
Mar 21 at 19:39
1
1
$begingroup$
With respect to which basis?
$endgroup$
– José Carlos Santos
Mar 21 at 18:37
$begingroup$
With respect to which basis?
$endgroup$
– José Carlos Santos
Mar 21 at 18:37
$begingroup$
I just did one of these a minute ago math.stackexchange.com/questions/3157130/…
$endgroup$
– Will Jagy
Mar 21 at 18:40
$begingroup$
I just did one of these a minute ago math.stackexchange.com/questions/3157130/…
$endgroup$
– Will Jagy
Mar 21 at 18:40
1
1
$begingroup$
The space of $2times2$ matrix has dimension $4$, so the matrix of $T$ must be $4times4$.
$endgroup$
– TheSilverDoe
Mar 21 at 18:41
$begingroup$
The space of $2times2$ matrix has dimension $4$, so the matrix of $T$ must be $4times4$.
$endgroup$
– TheSilverDoe
Mar 21 at 18:41
$begingroup$
but how could we multiply a 4*4 matrix with a 2*2 matrix?
$endgroup$
– Eric
Mar 21 at 19:39
$begingroup$
but how could we multiply a 4*4 matrix with a 2*2 matrix?
$endgroup$
– Eric
Mar 21 at 19:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your linear transformation is a map $T:M_2to M_2$, where $M_2$ is the four dimensional vector space of $2times 2$-matrices. A basis for this vector space is $E_11=beginpmatrix1&0\0&0endpmatrix$, $E_12=beginpmatrix0&1\0&0endpmatrix$, $E_21=beginpmatrix0&0\1&0endpmatrix$, and $E_22=beginpmatrix0&0\0&1endpmatrix$. Looking at your formula above, you should see that
$T(E_11)=E_22$, $T(E_12)=-E_12$, $T(E_21)=-E_21$, and $T(E_22)=E_11$. Therefore, the matrix for $T$ is this basis is
$$[T]=beginpmatrix0&0&0&1\0&-1&0&0\0&0&-1&0\1&0&0&0endpmatrix.$$
answered Mar 21 at 19:56
David HillDavid Hill
9,5361619
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your linear transformation is a map $T:M_2to M_2$, where $M_2$ is the four dimensional vector space of $2times 2$-matrices. A basis for this vector space is $E_11=beginpmatrix1&0\0&0endpmatrix$, $E_12=beginpmatrix0&1\0&0endpmatrix$, $E_21=beginpmatrix0&0\1&0endpmatrix$, and $E_22=beginpmatrix0&0\0&1endpmatrix$. Looking at your formula above, you should see that
$T(E_11)=E_22$, $T(E_12)=-E_12$, $T(E_21)=-E_21$, and $T(E_22)=E_11$. Therefore, the matrix for $T$ is this basis is
$$[T]=beginpmatrix0&0&0&1\0&-1&0&0\0&0&-1&0\1&0&0&0endpmatrix.$$
answered Mar 21 at 19:56
David HillDavid Hill
9,5361619
$endgroup$
add a comment |
$begingroup$
Your linear transformation is a map $T:M_2to M_2$, where $M_2$ is the four dimensional vector space of $2times 2$-matrices. A basis for this vector space is $E_11=beginpmatrix1&0\0&0endpmatrix$, $E_12=beginpmatrix0&1\0&0endpmatrix$, $E_21=beginpmatrix0&0\1&0endpmatrix$, and $E_22=beginpmatrix0&0\0&1endpmatrix$. Looking at your formula above, you should see that
$T(E_11)=E_22$, $T(E_12)=-E_12$, $T(E_21)=-E_21$, and $T(E_22)=E_11$. Therefore, the matrix for $T$ is this basis is
$$[T]=beginpmatrix0&0&0&1\0&-1&0&0\0&0&-1&0\1&0&0&0endpmatrix.$$
answered Mar 21 at 19:56
David HillDavid Hill
9,5361619
$endgroup$
add a comment |
$begingroup$
Your linear transformation is a map $T:M_2to M_2$, where $M_2$ is the four dimensional vector space of $2times 2$-matrices. A basis for this vector space is $E_11=beginpmatrix1&0\0&0endpmatrix$, $E_12=beginpmatrix0&1\0&0endpmatrix$, $E_21=beginpmatrix0&0\1&0endpmatrix$, and $E_22=beginpmatrix0&0\0&1endpmatrix$. Looking at your formula above, you should see that
$T(E_11)=E_22$, $T(E_12)=-E_12$, $T(E_21)=-E_21$, and $T(E_22)=E_11$. Therefore, the matrix for $T$ is this basis is
$$[T]=beginpmatrix0&0&0&1\0&-1&0&0\0&0&-1&0\1&0&0&0endpmatrix.$$
answered Mar 21 at 19:56
David HillDavid Hill
9,5361619
$endgroup$
Your linear transformation is a map $T:M_2to M_2$, where $M_2$ is the four dimensional vector space of $2times 2$-matrices. A basis for this vector space is $E_11=beginpmatrix1&0\0&0endpmatrix$, $E_12=beginpmatrix0&1\0&0endpmatrix$, $E_21=beginpmatrix0&0\1&0endpmatrix$, and $E_22=beginpmatrix0&0\0&1endpmatrix$. Looking at your formula above, you should see that
$T(E_11)=E_22$, $T(E_12)=-E_12$, $T(E_21)=-E_21$, and $T(E_22)=E_11$. Therefore, the matrix for $T$ is this basis is
$$[T]=beginpmatrix0&0&0&1\0&-1&0&0\0&0&-1&0\1&0&0&0endpmatrix.$$
answered Mar 21 at 19:56
David HillDavid Hill
9,5361619
answered Mar 21 at 19:56
David HillDavid Hill
9,5361619
answered Mar 21 at 19:56
David HillDavid Hill
9,5361619
answered Mar 21 at 19:56
David HillDavid Hill
9,5361619
9,5361619
add a comment |
add a comment |
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1
$begingroup$
With respect to which basis?
$endgroup$
– José Carlos Santos
Mar 21 at 18:37
$begingroup$
I just did one of these a minute ago math.stackexchange.com/questions/3157130/…
$endgroup$
– Will Jagy
Mar 21 at 18:40
1
$begingroup$
The space of $2times2$ matrix has dimension $4$, so the matrix of $T$ must be $4times4$.
$endgroup$
– TheSilverDoe
Mar 21 at 18:41
$begingroup$
but how could we multiply a 4*4 matrix with a 2*2 matrix?
$endgroup$
– Eric
Mar 21 at 19:39