# Find one basis of kernel, one basis of image of the linear transformation and it's defect

Find one basis of kernel, one basis of image of the linear transformation $T:\mathcal{M}_{2\times 2}(\mathbb{R})\rightarrow \mathcal{M}_{2\times 3}(\mathbb{R})$ and defect of $T$ where $T(X)=X \begin{bmatrix} 0 & 0 & 0 \\ 2 & -2 & 4 \\ \end{bmatrix}+3\cdot trace(X) \begin{bmatrix} 3 & -3 & 6 \\ -1 & 1 & -2 \\ \end{bmatrix}$

I don't know how to find the matrix of $T$. Is it $6\times 4$ matrix?

Could someone show how to find matrix of $T$?

Also, what is the defect of a linear transformation? I am familiar with the idea of defective matrix. Defective matrix is a square matrix that is not diagonalizable. What would be the defect of a matrix? What if a matrix is not squared (like in this case)?

The space of $2\times 2$ matrices has as basis the elementary matrices $e_{11},e_{12},e_{21},e_{22}$ with a $1$ in the $ij$ position and $0$ elsewhere, so you can calculate the matrix of $T$ relative to this basis and with the corresponding basis of the space of $2\times 3$ matrices. Note that $e_{ij}$ has trace $0$ if $i\neq j$ and trace $1$ if $i=j$, so that

\begin{align} T(e_{12})&=\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix} 0 & 0 & 0 \\ 2 & -2 & 4 \\ \end{pmatrix}+0 \begin{pmatrix} 3 & -3 & 6 \\ -1 & 1 & -2 \\ \end{pmatrix}=\begin{pmatrix}2&-2&4\\0&0&0\end{pmatrix}\\ T(e_{11})&=\begin{pmatrix}1&0\\0&0\end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 2 & -2 & 4 \\ \end{pmatrix}+3 \begin{pmatrix} 3 & -3 & 6 \\ -1 & 1 & -2 \\ \end{pmatrix}=\begin{pmatrix}9&-9&18\\-3&3&-6\end{pmatrix}\\ T(e_{21})&=\begin{pmatrix}0&0\\1&0\end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 2 & -2 & 4 \\ \end{pmatrix}+0 \begin{pmatrix} 3 & -3 & 6 \\ -1 & 1 & -2 \\ \end{pmatrix}=\begin{pmatrix}0&0&0\\0&0&0\end{pmatrix}\\ T(e_{22})&=\begin{pmatrix}0&0\\0&1\end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 2 & -2 & 4 \\ \end{pmatrix}+3 \begin{pmatrix} 3 & -3 & 6 \\ -1 & 1 & -2 \\ \end{pmatrix}=\begin{pmatrix}9&-9&18\\-1&1&-2\end{pmatrix}\end{align}

By looking at matrices as row vectors, say, you can consider $T$ as a transformation $T:\Bbb R^4\to\Bbb R^6$ using the isomorphisms that send canonical elementary matrices to canonical row vectors. If we use the order $e_{11},e_{12},e_{21},e_{22}$ and send it to $e_1,e_2,e_3,e_4$ and use the order $e_{11},e_{12},e_{13},e_{21},e_{22},e_{23}$ and send it to $e_1,\cdots,e_6$, we get the matrix $$\begin{pmatrix} 2&9&0&9\\-2 &-9&0&-9 \\4&18&0&18\\0&-3&0&-1\\0&3&0&1\\0&-6&0&-2\end{pmatrix}$$

Suppose that we have linear transformations $f:V\to W$ and $g:W\to Z$ for which $gf=0$, over finite dimensional vector spaces, and that $gf=0$. One then calls $$d=\dim W-\operatorname{rank}{f}-\operatorname{rank}{g}$$

the "defect". If one knows some homological algebra, $d$ is the dimension of the quotient space $\ker g/{\rm im}\, f$ which is the homology of $$V\longrightarrow W\longrightarrow Z$$ at $W$, since $\dim\ker g=\dim W-\operatorname{rank}g$ by the rank-nullity theorem.

• ♦ Thanks for detailed explanation. The matrix of $T$ is not squared, so what is it's defect? Commented Dec 24, 2015 at 18:23
• @user300044 I am not sure. I've added something, but it involves two linear transformations, not one.
– Pedro
Commented Dec 24, 2015 at 18:47