# Math Help - Do these matrices represent linear transformations?

1. ## Do these matrices represent linear transformations?

Hello,

I have a question to do which I'm a bit stuck on. Any and all help would be much appreciated!

Assume V is an inner product space. If dim(V) = 3, for which of the following matrices does there exist a self-adjoint linear transformation $T_{i}$ on V for which the matrix $A_{i}$ represents $T_{i}$ with respect to some basis?

$A_{0}=\begin{pmatrix}
2 & 0 & 0\\
0 & 1 & 1\\
0 & 0 & 0
\end{pmatrix}
A_{1}=\begin{pmatrix}
2 & 0 & 0\\
0 & 1 & 1\\
0 & 0 & 1
\end{pmatrix}
A_{2}=\begin{pmatrix}
2 & 0 & 0\\
0 & 1 & 1\\
0 & 0 & 2
\end{pmatrix}$

For the first one I'm thinking there's not, because of the row of 0's at the bottom. However, I'm still not 100% sure.

(I understand what a self-adjoint linear transformation is, and I also know that the eigenvectors of V will form an orthonormal basis of V.)

2. Originally Posted by amy99
Hello,

I have a question to do which I'm a bit stuck on. Any and all help would be much appreciated!

Assume V is an inner product space. If dim(V) = 3, for which of the following matrices does there exist a self-adjoint linear transformation $T_{i}$ on V for which the matrix $A_{i}$ represents $T_{i}$ with respect to some basis?

$A_{0}=\begin{pmatrix}
2 & 0 & 0\\
0 & 1 & 1\\
0 & 0 & 0
\end{pmatrix}
A_{1}=\begin{pmatrix}
2 & 0 & 0\\
0 & 1 & 1\\
0 & 0 & 1
\end{pmatrix}
A_{2}=\begin{pmatrix}
2 & 0 & 0\\
0 & 1 & 1\\
0 & 0 & 2
\end{pmatrix}$

For the first one I'm thinking there's not, because of the row of 0's at the bottom. However, I'm still not 100% sure.

(I understand what a self-adjoint linear transformation is, and I also know that the eigenvectors of V will form an orthonormal basis of V.)
Eigenvectors I obtained for $A_0$
$\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}, \begin{bmatrix}
0\\
1\\
0
\end{bmatrix}, \begin{bmatrix}
0\\
-1\\
1
\end{bmatrix}$

Eigenvectors for $A_1$
$\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}, \begin{bmatrix}
0\\
1\\
0
\end{bmatrix}$

$A_3$
$\begin{bmatrix}
1\\
0\\
0
\end{bmatrix}, \begin{bmatrix}
0\\
1\\
1
\end{bmatrix}, \begin{bmatrix}
0\\
1\\
0
\end{bmatrix}$

3. Thank you, dwsmith. From what I can tell, then, $A_{1}$ cannot represent a linear transformation, because it only has two eigenvectors, and so cannot form an orthonormal basis of V (2 vectors cannot span V). Does it remain, then, to check whether the eigenvectors of $A_{0}$ and $A_{2}$ form an orthonormal set?

4. Originally Posted by amy99
Thank you, dwsmith. From what I can tell, then, $A_{1}$ cannot represent a linear transformation, because it only has two eigenvectors, and so cannot form an orthonormal basis of V (2 vectors cannot span V). Does it remain, then, to check whether the eigenvectors of $A_{0}$ and $A_{2}$ form an orthonormal set?
Be careful, here, every matrix can represent a linear transformation. But in your post you said self-adjoint linear transformation. Every self-adjoint linear transformation has a "complete set" of eigenvectors- a set of independent eigenvectors equal to the size of the matrix. That means that the eigenvectors could be used as a basis for the space.

You also said " Does it remain, then, to check whether the eigenvectors of $A_0$ and $A_2$ form an orthonormal set? "

An "orthonormal" set consists of vectors that all have length 1 and are orthogonal. In fact, here, both sets contain a vector that does not have length 1 so these are NOT "orthonormal". All you need are that the vectors are independent.