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Suppose $T(M) = \left( \begin{array}{ccc} 1 & 1 \\ 1 & 1 \end{array} \right)M-M\left( \begin{array}{ccc} 1 & 1 \\ 1 & 1 \end{array} \right)$ is a linear transformation from $\mathbb{R}^{2\times 2}$ to $\mathbb{R}^{2\times 2}$ with respect to the basis $\mathcal{B} = \left( \begin{array}{ccc} 1 & 1 \\ -1 & -1 \end{array} \right),\left( \begin{array}{ccc} 1 & -1 \\ 1 & -1 \end{array} \right),\left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right),\left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \end{array} \right)$

I'm told that the $\mathcal{B}$ matrix of the transformation is $\left( \begin{array}{cccc} 2 & 0&0&0 \\ 0 & -2&0&0\\ 0&0&0&0&\\ 0&0&0&0 \end{array} \right)$, and I'm asked to find a basis for the kernel and image of $T$.

Can someone please explain how we can do this? I know a basis for the $im(T)$ would be given by the linearly independent columns in $\mathcal{B}$, but what are they in terms of $\mathbb{R}^{2\times 2}$?

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  • $\begingroup$ Each column in the $4 \times 4$ matrix corresponds to one of the four basis elements. $\endgroup$ – copper.hat Nov 19 '14 at 20:30
  • $\begingroup$ Is this true for any $\mathcal{B}$ matrix for a transformation with some basis? $\endgroup$ – Parker Nov 19 '14 at 20:39
  • $\begingroup$ With that thought, I can see that a basis for the image of $T$ will be $2M_1, -2M_2$ (where $M_1,M_2$ are the first two basis elements), but how can we find the kernel of $T$? $\endgroup$ – Parker Nov 19 '14 at 20:40
  • $\begingroup$ If I understand you correctly, then yes. Given an ordered basis, there is a matrix corresponding to the transformation and the columns will correspond to the basis elements. $\endgroup$ – copper.hat Nov 19 '14 at 20:41
  • $\begingroup$ What is the kernel of the $4\times 4$ matrix? $\endgroup$ – copper.hat Nov 19 '14 at 20:41
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Not an answer as such, but another way of looking at the problem:

Note that the symmetric $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ has an eigenvalue $\lambda_1=2$ corresponding to the eigenvector $v_1={1 \over \sqrt{2}} (1,1)^T$, and an eigenvalue $\lambda_2 = 0$ corresponding to the eigenvector $v_1={1 \over \sqrt{2}} (1,-1)^T$.

Note that the matrices $v_i v_j^T$ are a basis for the space of $2 \times 2$ matrices, and $T(v_i v_j^T) = (\lambda_i - \lambda_j)v_i v_j^T$. Hence $T$ is diagonal in the basis $v_i v_j^T$, and we see that $\ker T = \operatorname{sp} \{ v_1 v_1^T, v_2 v_2^T \}$, and ${\cal R} T = \operatorname{sp} \{ v_1 v_2^T, v_2 v_1^T \}$.

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