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

$$A= \begin{pmatrix} 5 & -2 & 4\\ 0 & 3 & t\\ 0 & 0 & 5 \end{pmatrix}$$

The Attempt at a Solution

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

$$det(A-xI)=det \begin{pmatrix} 5-x & -2 & 4\\ 0 & 3-x & t\\ 0 & 0 & 5-x \end{pmatrix}$$
$$=(5-x)det \begin{pmatrix} 3-x & t\\ 0 & 5-x \end{pmatrix}$$
$$= (5-x)((3-x)(5-x)-0t)$$

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

$$A= \begin{pmatrix} 5 & -2 & 4\\ 0 & 3 & t\\ 0 & 0 & 5 \end{pmatrix}$$

The Attempt at a Solution

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

$$det(A-xI)=det \begin{pmatrix} 5-x & -2 & 4\\ 0 & 3-x & t\\ 0 & 0 & 5-x \end{pmatrix}$$
$$=(5-x)det \begin{pmatrix} 3-x & t\\ 0 & 5-x \end{pmatrix}$$
$$= (5-x)((3-x)(5-x)-0t)$$

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 [itex]\begin{pmatrix}x \\ y \\ z\end{pmatrix}[/tex] is an eigenvector for A, with eigenvalue 5, then
$$\begin{pmatrix}5 & -2 & 4 \\ 0 & 3 & t \\ 0 & 0 & 5\end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= 5\begin{pmatrix}x \\ y \\ z\end{pmatrix}$$
$$\begin{pmatrix}5x- 2y+ 4z \\ 3y+ tz \\ 5z\end{pmatrix}= \begin{pmatrix}5x \\ 5y \\ 5z\end{pmatrix}$$

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.

 

[PirateFan308]

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

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 $$\begin{pmatrix}x' \\ y' \\ z'\end{pmatrix}$$ is also an eigenvector for A

$$\begin{pmatrix} 5&-2&4\\0&3&t\\0&0&5\end{pmatrix}\begin{pmatrix}x'\\y'\\z'\end{pmatrix} = 3\begin{pmatrix}x'\\y'\\z'\end{pmatrix}$$

Where there are no free variables, no matter what t is, and $$5x'-2y'+4z'=3x'$$ and $$3y'+tz'=3z'$$ and $$5z'=3z'$$ means that $$x'=y'=z'=0$$ 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?