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Math Help - Do these matrices represent linear transformations?

  1. #1
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    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}<br />
2 & 0 & 0\\ <br />
0 & 1 & 1\\ <br />
0 & 0 & 0<br />
\end{pmatrix}<br />
A_{1}=\begin{pmatrix}<br />
2 & 0 & 0\\ <br />
0 & 1 & 1\\ <br />
0 & 0 & 1<br />
\end{pmatrix}<br />
A_{2}=\begin{pmatrix}<br />
2 & 0 & 0\\ <br />
0 & 1 & 1\\ <br />
0 & 0 & 2<br />
\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.)
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  2. #2
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    Quote Originally Posted by amy99 View Post
    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}<br />
2 & 0 & 0\\ <br />
0 & 1 & 1\\ <br />
0 & 0 & 0<br />
\end{pmatrix}<br />
A_{1}=\begin{pmatrix}<br />
2 & 0 & 0\\ <br />
0 & 1 & 1\\ <br />
0 & 0 & 1<br />
\end{pmatrix}<br />
A_{2}=\begin{pmatrix}<br />
2 & 0 & 0\\ <br />
0 & 1 & 1\\ <br />
0 & 0 & 2<br />
\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}<br />
1\\ <br />
0\\ <br />
0<br />
\end{bmatrix}, \begin{bmatrix}<br />
0\\ <br />
1\\ <br />
0<br />
\end{bmatrix}, \begin{bmatrix}<br />
0\\ <br />
-1\\ <br />
1<br />
\end{bmatrix}

    Eigenvectors for A_1
    \begin{bmatrix}<br />
1\\ <br />
0\\ <br />
0<br />
\end{bmatrix}, \begin{bmatrix}<br />
0\\ <br />
1\\ <br />
0<br />
\end{bmatrix}

    A_3
    \begin{bmatrix}<br />
1\\ <br />
0\\ <br />
0<br />
\end{bmatrix}, \begin{bmatrix}<br />
0\\ <br />
1\\ <br />
1<br />
\end{bmatrix}, \begin{bmatrix}<br />
0\\ <br />
1\\ <br />
0<br />
\end{bmatrix}
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  3. #3
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    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?
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  4. #4
    MHF Contributor

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    Quote Originally Posted by amy99 View Post
    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.
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