Find the matrix of the given linear transformation T with respect to the given basis.
Determine whether T is an isomorphism. If T isn't an isomorphism find bases of the kernel and image of T, and thus determine the rank of T.
T(M) = M$\begin{bmatrix}1&2\\0&1\end{bmatrix}$ – $\begin{bmatrix}1&2\\0&1\end{bmatrix}$M from $U^{2×2}$ to $U^{2×2}$
with respect to the standard basis $\mathfrak A$ $\text = (1, t, t^2)$
Which is the basis for all second degree polynomials.
I know that T is not an isomorphism because $\begin{bmatrix}1&0\\0&1\end{bmatrix}$, $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ are in the kernel.
And I have found A = $\begin{bmatrix}0&0&0\\2&0&-2\\0&0&0\end{bmatrix}$
I'm unsure how to find the image and rank of T. There is a similar problem posted. But it has different bases.
Best Answer
I'm assuming that $U^{2\times 2}$ is the space of $2\times 2$ matrices. Your map $T:U^{2\times 2}\to U^{2\times 2}$ is $T(X)=AX-XA$ where $$ A = \left[\begin{array}{rr} 1 & 2 \\ 0 & 1 \end{array}\right] $$ It sounds like you're trying to represent $T$ as a matrix. To do so, we need a basis for $U^{2\times 2}$. The standard basis is \begin{align*} E_{11} &= \left[\begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array}\right] & E_{12} &= \left[\begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array}\right] & E_{21} &= \left[\begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array}\right] & E_{22} &= \left[\begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array}\right] \end{align*} Note that \begin{array}{rcrcrcrcrcrcrcrcrcrc} T(E_{11}) &=& \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix} & = & 0\cdot E_{11} &+& 2\cdot E_{12} &+& 0\cdot E_{21} &+& 0\cdot E_{22} \\ T(E_{12}) &=& \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} &=& 0\cdot E_{11} &+& 0\cdot E_{12} &+& 0\cdot E_{21} &+& 0\cdot E_{22} \\ T(E_{21}) &=& \begin{bmatrix} -2 & 0 \\ 0 & 2 \end{bmatrix} &=& -2\cdot E_{11} &+& 0\cdot E_{12} &+& 0\cdot E_{21} &+& 2\cdot E_{22} \\ T(E_{22}) &=& \begin{bmatrix} 0 & -2 \\ 0 & 0 \end{bmatrix} &=& 0\cdot E_{11} &+& -2\cdot E_{12} &+& 0\cdot E_{21} &+& 0\cdot E_{22} \end{array} This shows that our matrix representation of $T$ is $$ [T]= \left[\begin{array}{rrrr} 0 & 0 & -2 & 0 \\ 2 & 0 & 0 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 \end{array}\right] $$ Now, use your favorite method to compute bases for the null space and column space of $[T]$.
The vectors \begin{align*} \langle 1,0,0,1\rangle && \langle0,1,0,0\rangle \end{align*} form a basis for the null space of $[T]$. This tells us that the matrices \begin{array}{rcrcrcrcrcrcrcrcrcrc} 1\cdot E_{11} &+& 0\cdot E_{12} &+& 0\cdot E_{21} &+& 1\cdot E_{22} &=& \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \\ 0\cdot E_{11} &+& 1\cdot E_{12} &+& 0\cdot E_{21} &+& 0\cdot E_{22} &=& \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} \\ \end{array} form a basis for the kernel of $T$.
Can you find a basis for the column space of $[T]$ and use this to find a basis for the image of $T$?