# Find Transformation Matrix $T$ relative to new bases such that $T$ is in diagonal form

$T$ is a linear transformation from $R^2 \rightarrow R^3$. The matrix of $T$ = $\left[\begin{array}{cc}1 & -1 \\0 & 0 \\1 & 1\end{array}\right]$. Question: how to find bases $(e_1,e_2)$ in $R^2$ and $(w_1,w_2,w_3)$ in $R^3$ such that $T$ is in diagonal form.

I started by making $w_1 = (1,0,1)^T$, $w_2 = (-1,0,1)^T$, and $w_3 = (0,1,0)^T$. So far, I know $T(e_1) = w_1$. $T(e_2) = w_2$.

Is this all I need to do? I'm still still a little shaky on how to set up and solve this problem.

Side note: I know of an answer: $e_1=i,e_2=i+j,w_1=(1,0,1)^T,w_2=(0,0,2)^T, w_3=(0,1,0)^T$. But still not sure why this is correct. How do you arrive at this answer?

I need someone to show how to walk through the steps of solving this without using SVD.

• The concept of a diagonal matrix usually refers to square matrices. What definition are you using here? – Reveillark Mar 22 '15 at 3:10
• @Reveillark, that is very diplomatic. – abel Mar 22 '15 at 3:15
• en.wikipedia.org/wiki/Main_diagonal. A diagonal matrix is one where the main diagonal of a matrix is non-zero. All other entries are 0. – larry Mar 22 '15 at 3:19
• SDV (Singular Value Decompositon): $T=U\Sigma V^T$, where $\Sigma=\left[\begin{array}{cc} \sigma_1&0\\0&\sigma_2\\0&0\end{array}\right]$, and $U\in\mathbb{R}^{3\times 3}$, $V\in\mathbb{R}^{2\times 2}$ are orthogonal matrices. – xecafe Mar 22 '15 at 4:37
• @xecafe (a) I think you mean SVD (not SDV), and (b) how is SVD supposed to help? – larry Mar 22 '15 at 13:35

Let $B=\{v_1,v_2\}$ be any basis for $\mathbb{R^2},\;\;$ where $v_1=\begin{bmatrix} a \\ b\end{bmatrix}$ and $v_2=\begin{bmatrix} c\\d\end{bmatrix}$.

We need to find a basis $B^{\prime}=\{w_1, w_2, w_3\}$ for $\mathbb{R^3}$ such that

$\hspace{.3 in}T(v_1)=k w_1$ and $T(v_2)=l w_2$ for some $k, l\ne0$, so

let $w_1=k^{-1}(T(v_1))=k^{-1}\begin{bmatrix}a-b\\0\\a+b\end{bmatrix}$ and $w_2=l^{-1}(T(v_2))=l^{-1}\begin{bmatrix}c-d\\0\\c+d\end{bmatrix}$ for any $k,l\ne0$.

If $w_3=\begin{bmatrix}e\\f\\g\end{bmatrix}$ where $f\ne0$ and $B^{\prime}=\{w_1,w_2,w_3\}$,

then the matrix for T with respect to $B$ and $B^{\prime}$ is given by

$\displaystyle[T]_{B,B^{\prime}}=\left[[T(v_1)]_{B^{\prime}}\lvert[T(v_2)]_{B^{\prime}}\right]=\begin{bmatrix}k&0\\0&l\\0&0\end{bmatrix}\;\;$ since $T(v_1)=kw_1$ and $T(v_2)=lw_2$.

In your example, $v_1=\vec{i}, v_2=\vec{i}+\vec{j}, \;k=l=f=1, \;e=g=0$.

As another example, take $v_1=\begin{bmatrix}4\\1\end{bmatrix}$ and $v_2=\begin{bmatrix}1\\3\end{bmatrix}$, and let $k=5, l=8, f=6, e=g=1$.

Then $w_1=\begin{bmatrix}\frac{3}{5}\\0\\1\end{bmatrix}$, $w_2=\begin{bmatrix}-\frac{1}{4}\\0\\\frac{1}{2}\end{bmatrix}$, $w_3=\begin{bmatrix}1\\6\\1\end{bmatrix}$, and $\displaystyle[T]_{B,B^{\prime}}=\begin{bmatrix}5&0\\0&8\\0&0\end{bmatrix}\;\;$.

Compute the singular value decomposition (SVD) of $T$, i.e. $T=UWV^*$ where $W$ is diagonal. Then the columns of $U$ will yield the basis for $\mathbb R^3$ that you seek, and the columns of $V$ will yield the basis of $\mathbb R^2$ that you seek.

Here, you get $$T=UWV^*=\begin{bmatrix} -1/\sqrt{2} & 1/\sqrt{2} & 0\\ 0 & 0 & 1\\ 1/\sqrt{2} & 1/\sqrt{2} & 0\end{bmatrix} \begin{bmatrix} \sqrt 2 & 0\\ 0 & \sqrt 2\\ 0 & 0\end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}^*,$$ so the basis for $\mathbb R^3$ is the columns of $U$ while the basis for $\mathbb R^2$ is the columns of $V$.

• is there a way to do this without using SVD? – larry Apr 1 '15 at 23:20
• can you show a way to do this without using SVD? – larry Apr 2 '15 at 2:31
• This is the only approach I know of off the top of my head. – JohnD Apr 3 '15 at 2:57

I think, one should keep in mind, that it is not appropriate to speak of THE matrix of a linear transformation $T: V \rightarrow W$ between finite dimensional vector spaces $V$ and $W$ over the field $K$, unless one has chosen bases for both spaces BEFOREHAND. If we speak of THE matrix of $T$ without further elaboration, we imply to already have chosen the standard basis in $V$ and $W$ (if such a thing as a standard basis is obvious at all).

But having consciously chosen a BASIS ($v_1, \ldots, v_n$) in $V$ and ($w_1, \ldots, w_m$) in $W$, there is a theorem (somewhere in any linear algebra text I want to know), that for every linear transformation $T$ there is EXACTLY ONE matrix $A_T$ with entries $a_{ij}$ in $K$, which satisfies

$$T(v_i) \ = \ \sum_{i = 1}^m a_{ij}w_j \, .$$

So if one has chosen bases ($v_1, v_2$) for $\mathbb{R}^2$ and ($w_1, w_2, w_3$) for $\mathbb{R}^3$, as you did, one can be sure, that if $T(v_1) \ = \ w_1$ and $T(v_2) \ = \ w_2$, then the matrix $A_T$ will be $$\begin{pmatrix} 1 \ 0 \\ 0 \ 1 \\ 0 \ 0 \end{pmatrix} \, ,$$ because this matrix maps the vectors in the basis of $V$ correctly, and being unique, there can be no other one to do the job. Note, that the arrangement of the basis vectors is relevant here. If $T$ is as above and your basis in $\mathbb{R}^2$ is ($v_1,v_2$), while your basis in $\mathbb{R}^3$ is ($w_1,w_3,w_2$), you end up with $$\begin{pmatrix} 1 \ 0 \\ 0 \ 0 \\ 0 \ 1 \end{pmatrix} \, .$$ If you had chosen ($v_1, v_2$) and ($w_1, w_2, w_3$) as BASES, such that $T(v_1) \ = \ 2 w_1$ and $T(v_2) \ = \ w_3$, for your $T$ for example ($(1,0)^t, (0,1)^t$) and ($(1/2,0,1/2)^t, (0,1,1)^t, (-1,0,1)^t$), then you would have ended up with $$\begin{pmatrix} 2 \ 0 \\ 0 \ 0 \\ 0 \ 1 \end{pmatrix} \, .$$

The SVD will always succeed, but with some deliberation it is really easy to write down any number of diagonal matrices for any linear transformation by choosing the bases in $V$ and $W$ appropriately.