# Find a real number $t$ such that $\alpha$ is represented with respect to the basis

Doing some self study as this topic really stumps me, every time I see it I cringe. I believe this question requires knowledge of change of basis and using $\phi_{BD}(\alpha)$.

Question: Let $\alpha\in End(\Bbb R^3)$ be represented with respect to the canonical basis by the matrix $\begin{bmatrix} 4 & 4 & 1 \\ 2 & 2 & 3 \\ 2 & 2 & 3 \\ \end{bmatrix}$. Find a real number $t$ such that $\alpha$ is represented with respect to the basis $B=\begin{Bmatrix}\begin{bmatrix} t \\-2\\0 \end{bmatrix} \begin{bmatrix} -1 \\t \\2 \\\end{bmatrix}\begin{bmatrix} 2\\2 \\t \\ \end{bmatrix}\end{Bmatrix}$ by the matrix $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \\ \end{bmatrix}$.

Can anyone explain this question? I know what the canonical basis is but im not seeing how to connect everything in this problem. Do you work backwards from the first matrix to find $\alpha$ then use alpha to find t?

I greatly appreciate someone explaining this to me.

Edited to add correct matrices: (I am leaving the original so the comments and answer given still make sense)

Let $\alpha\in End(\Bbb R^3)$ be represented with respect to the canonical basis by the matrix $\begin{bmatrix} 2 & 2 & 0 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \\ \end{bmatrix}$. Find a real number $t$ such that $\alpha$ is represented with respect to the basis $B=\begin{Bmatrix}\begin{bmatrix} t \\-1\\0 \end{bmatrix} \begin{bmatrix} -2 \\t \\1 \\\end{bmatrix}\begin{bmatrix} 1\\1 \\t \\ \end{bmatrix}\end{Bmatrix}$ by the matrix $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \\ \end{bmatrix}$.

• It probably means given the matrix $T$ which is built with the three columns in B ( in some order ), find t so that $T^{-1}\text{diag}([0,1,4])T = [4,4,1\cdots]$ – mathreadler Oct 17 '16 at 17:32
• Are you sure the example is like that? It looks to me like $t$ doesn't exist in this concrete example. – Martin Argerami Oct 17 '16 at 17:59
• @MartinArgerami nope, looks like I have half from one example and half from another....no idea how i did that. I cant fix it on my phone at the moment but I'm more looking for the process. – Aksel'sRose Oct 17 '16 at 18:04

Because the matrix is diagonal, the form is very simple. If $B=\{b_1,b_2,b_3\}$, the matrix $$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \\ \end{bmatrix}$$ means that $$\tag{*} \alpha b_1=0,\ \ \alpha b_2=b_2,\ \ \alpha b_3=4b_3.$$ So $$\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} =\alpha b_1=\begin{bmatrix} 4 & 4 & 1 \\ 2 & 2 & 3 \\ 2 & 2 & 3 \\ \end{bmatrix} \begin{bmatrix} t \\ -2\\ 0\\ \end{bmatrix} =\begin{bmatrix} 4t-8 \\ 2t-4 \\ 2t-4 \\ \end{bmatrix}.$$ So you need $t=2$. Now you still need to check that this $t$ also works for $b_2$ and $b_3$, which doesn't in this case.
In general, that $A$ is the matrix for the basis $b_1,b_2,b_3$ means that $$\alpha b_1=A_{11}b_1+A_{12}b_2+A_{13}b_3,$$ and so on.
Edit: with the new data, the equations in $(*)$ now are $$\begin{bmatrix} 2 & 2 & 0 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \\ \end{bmatrix}\begin{bmatrix} t \\-1\\0 \end{bmatrix} =\begin{bmatrix} 2t-2 \\t-1\\t-1 \end{bmatrix},$$ which forces $t=1$. Now, for the other two vectors, $$\begin{bmatrix} 2 & 2 & 0 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \\ \end{bmatrix} \begin{bmatrix} -2 \\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \\ 1 \\ \end{bmatrix}$$ and $$\begin{bmatrix} 2 & 2 & 0 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 4 \\ 4 \\ 4 \\ \end{bmatrix}=4\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$$
• It does get the only viable $t$. – Martin Argerami Oct 17 '16 at 19:38