# Basis of kernel linear transformation

For this matrix $A$ acting on $V = \mathbb{R}^3$ \begin{pmatrix} -1 & 1 & 0\\ 1 & 0 & 1\\ 1 & 0 & 0 \\ \end{pmatrix}

Find a basis for $\ker(A + I)^2$ and for $\ker(A - I)$ and who that these subspaces are $T_A$-Invariant

Show that $V$ is the direct sum of the above two $T_A$-Invariant subspaces.

Display the matrix of $T_A$ relative to this new basis.

Could anyone walk me through this question? It's a sample question for my upcoming midterm. I'm not too sure about showing subspaces of $T_A$-invariant.

Thanks

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How does $A$ act on $\mathbb R^3$? – Diego Silvera Feb 17 '13 at 0:30
Can you at least do the first steps? Are you able to find the appropriate basis vectors and to Jordanize the matrix? – EuYu Feb 17 '13 at 0:31
What does TA-invariant mean? Did you mean A-invariant? – 1015 Feb 17 '13 at 1:12
Sorry I meant $T_A$ Invariant where it is a linear transformation. I can find the basis vectors. Is it necessary to jordanize it? – Yam Feb 17 '13 at 1:16
By $T_A$, do you mean the linear map $T_A(x)=Ax$? – Avi Steiner Feb 17 '13 at 1:26

$A+I=\pmatrix{0&1&0\\1&1&1\\1&0&1}$. Now since a matrix with columns $[v_1\ v_2\ \dots]$ multiplied (from right) by a vector $\pmatrix{a_1\\a_2\\ \vdots}$ yields $a_1v_1+a_2v_2+\dots$, it is easily seen that already for $v_1:=\pmatrix{1\\0\\-1}$ $$(A+I)v_1=0\, ,$$ so $v_1\in\ker(A+I)\subseteq\ker(A+I)^2$.
Now $(A+I)^2=\pmatrix{1&1&1\\2&2&2\\1&1&1}$. Now all columns are the same, and besides $v_1$, we easily find that $v_2:=\pmatrix{1\\-1\\0}$ also satisfies $(A+I)^2v_2=0$, and is clearly independent from $v_1$. The image of $(A+I)^2$ contains exactly the multiples of $\pmatrix{1\\2\\1}$, hence it has dimension $1$, so $\dim\ker(A+I)^2=2$, and we already found a basis: $v_1,v_2$.
For $A$-invariance, calculate $Av_1$ and $Av_2$. We get $Av_1=-v_1$ (also because $(A+I)v_1=0$), and $Av_2=-v_1-v_2$. So, this subspace $\ker(A+I)^2=\langle v_1,v_2\rangle$ is invariant under the action of $A$.
$A-I=\pmatrix{-2&1&0\\1&-1&1\\1&0&-1}$. Now, calculating by columns, for getting $0$ as their combination, add the first column to the third, plus twice the middle one: $$v_3:=\pmatrix{1\\2\\1}, \quad (A-I)v_3=0$$ $v_3$ is independent from $v_1$ and $v_2$ because, either you can check that $(A+I)^2v_3\ne 0$, or we can argue that that the coordinates of all vectors in $\langle v_1,v_2\rangle$ sum up to $0$, so $v_3\notin\langle v_1,v_2\rangle$. Then, the direct composition of all the 3d space is given: $$\Bbb R^3=\langle v_1,v_2\rangle \oplus\langle v_3\rangle$$ We also need $Av_3$ for its matrix in the new basis $v_1,v_2,v_3$, but it is just $v_3$, because $(A-I)v_3=0$. So, its matrix in the new basis is: $$\pmatrix{-1&-1&0\\0&-1&0\\0&0&1}$$ (the middle column represents that $Av_2=-v_1-v_2$).
For completeness, let me add that if $B=\left[v_1\ v_2\ v_3\right]$, then this (almost Jordan formed) matrix in the new basis is just $B^{-1}AB$.