Linear Algebra - diagonalizable matrix

[PirateFan308]

Homework Statement

For the following matrix, find the value of t, if any, so that the following matrix is diagonalizable

A=<br /> \begin{pmatrix}<br /> 5 &amp; -2 &amp; 4\\<br /> 0 &amp; 3 &amp; t\\<br /> 0 &amp; 0 &amp; 5<br /> \end{pmatrix}<br />

The Attempt at a Solution

In order for A to be diagonalizable, we need 3 linearly independent eigenvectors, that is, 3 linearly independent eigenvalues

det(A-xI)=det<br /> \begin{pmatrix}<br /> 5-x &amp; -2 &amp; 4\\<br /> 0 &amp; 3-x &amp; t\\<br /> 0 &amp; 0 &amp; 5-x<br /> \end{pmatrix}<br />
=(5-x)det<br /> \begin{pmatrix}<br /> 3-x &amp; t\\<br /> 0 &amp; 5-x<br /> \end{pmatrix}<br />
= (5-x)((3-x)(5-x)-0t)

The eigenvalues are 3 and 5.
Obviously, it doesn't matter what t is, we will not be able to get the matrix A to be diagonalizable.

My professor said that he thought there was one correct value for t (but he wasn't sure). Is what I've done correct?

 

[Ray Vickson]

Homework Statement

For the following matrix, find the value of t, if any, so that the following matrix is diagonalizable

A=<br /> \begin{pmatrix}<br /> 5 &amp; -2 &amp; 4\\<br /> 0 &amp; 3 &amp; t\\<br /> 0 &amp; 0 &amp; 5<br /> \end{pmatrix}<br />

The Attempt at a Solution

In order for A to be diagonalizable, we need 3 linearly independent eigenvectors, that is, 3 linearly independent eigenvalues

det(A-xI)=det<br /> \begin{pmatrix}<br /> 5-x &amp; -2 &amp; 4\\<br /> 0 &amp; 3-x &amp; t\\<br /> 0 &amp; 0 &amp; 5-x<br /> \end{pmatrix}<br />
=(5-x)det<br /> \begin{pmatrix}<br /> 3-x &amp; t\\<br /> 0 &amp; 5-x<br /> \end{pmatrix}<br />
= (5-x)((3-x)(5-x)-0t)

The eigenvalues are 3 and 5.
Obviously, it doesn't matter what t is, we will not be able to get the matrix A to be diagonalizable.

My professor said that he thought there was one correct value for t (but he wasn't sure). Is what I've done correct?

It is not a matter of whether or not you have a repeated eigenvalue; what is important is whether the geometric and algebraic multiplicities of an eigenvalue are the same. In other words, the dimensionality of the eigenspace for eigenvalue λ=5 needs to be determined. If λ=5 has two linearly-independent eigenvectors, then the matrix will be diagonalizable, so you need to find eigenvectors.

RGV

 

[deluks917]

There is a value of t for which the matrix is diagonalizable.

 

[HallsofIvy]

If \begin{pmatrix}x \\ y \\ z\end{pmatrix}[/tex] is an eigenvector for A, with eigenvalue 5, then<br /> \begin{pmatrix}5 &amp;amp; -2 &amp;amp; 4 \\ 0 &amp;amp; 3 &amp;amp; t \\ 0 &amp;amp; 0 &amp;amp; 5\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= 5\begin{pmatrix}x \\ y \\ z\end{pmatrix}<br /> \begin{pmatrix}5x- 2y+ 4z \\ 3y+ tz \\ 5z\end{pmatrix}= \begin{pmatrix}5x \\ 5y \\ 5z\end{pmatrix}<br /> <br /> and so must satify the equations 5x- 2y+ z= 5x, 3y+ tz= 5y, 5z= 5z.<br /> <br /> Note that the &quot;5x&quot; terms cancel in the first equation and then there is NO &quot;x&quot; in the equations. One eigenvector with eigenvalue is &lt;1, 0, 0&gt;. There is a single value of t that makes the first two equations identical, giving more than one dimension for the eigenspace.

 

[PirateFan308]

If \begin{pmatrix}x \\ y \\ z\end{pmatrix} is an eigenvector for A, with eigenvalue 5, then
\begin{pmatrix}5 &amp; -2 &amp; 4 \\ 0 &amp; 3 &amp; t \\ 0 &amp; 0 &amp; 5\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= 5\begin{pmatrix}x \\ y \\ z\end{pmatrix}
\begin{pmatrix}5x- 2y+ 4z \\ 3y+ tz \\ 5z\end{pmatrix}= \begin{pmatrix}5x \\ 5y \\ 5z\end{pmatrix}

and so must satify the equations 5x- 2y+ z= 5x, 3y+ tz= 5y, 5z= 5z.

Note that the "5x" terms cancel in the first equation and then there is NO "x" in the equations. One eigenvector with eigenvalue is <1, 0, 0>. There is a single value of t that makes the first two equations identical, giving more than one dimension for the eigenspace.

So for eigenvalue 3, if \begin{pmatrix}x&#039; \\ y&#039; \\ z&#039;\end{pmatrix} is also an eigenvector for A

\begin{pmatrix} 5&amp;-2&amp;4\\0&amp;3&amp;t\\0&amp;0&amp;5\end{pmatrix}\begin{pmatrix}x&#039;\\y&#039;\\z&#039;\end{pmatrix} = 3\begin{pmatrix}x&#039;\\y&#039;\\z&#039;\end{pmatrix}

Where there are no free variables, no matter what t is, and 5x&#039;-2y&#039;+4z&#039;=3x&#039; and 3y&#039;+tz&#039;=3z&#039; and 5z&#039;=3z&#039; means that x&#039;=y&#039;=z&#039;=0 correct?

So if t=4, the matrix A will always be diagonalizable?

We can say that the 3 linearly independent eigenvectors are (if for the second eigenvector, we take z=1 and x=1 and for the third eigenvector we take z=2 and x=2 because both are free):
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} ~and~~ \begin{pmatrix}1 \\ 2 \\ 1 \end{pmatrix} ~and~~ \begin{pmatrix}1 \\ 4 \\ 2 \end{pmatrix}

But there are infinitely many eigenvectors, I just chose these 3. Is this correct?

 

[PirateFan308]

But i just realized that the eigenvector \begin{pmatrix} 0\\0\\0 \end{pmatrix}
isn't a proper eigenvector because it is all 0. So I could take the eigenvectors (if for the first eigenvector, we take z=3 and x=1):
\begin{pmatrix} 1 \\ 6 \\ 3 \end{pmatrix} ~and~~ \begin{pmatrix}1 \\ 2 \\ 1 \end{pmatrix} ~and~~ \begin{pmatrix}1 \\ 4 \\ 2 \end{pmatrix}

Is this now correct?