Finding the Eigenvalue of a Matrix

In summary: Homework Statement Find the eigenvalues of the matrix ##\begin{bmatrix}4 & 0 & 0 \\0 & 0 & 0 \\1 & 0 & -3\end{bmatrix}##Homework Equations##Ax=λx##The Attempt at a SolutionI'm having some trouble finding the eigenvalues of this matrix.The eigenvalue of a matrix is a scalar λ such that ##Ax=λx##.To find the eigenvalue, I tried the following:##\begin{bmatrix}4 & 0 & 0 \\0 & 0 & 0 \\1 & 0 & -3\end{bmatrix} \begin{b
  • #1
Drakkith
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Homework Statement


Find the eigenvalues of the matrix ##\begin{bmatrix}
4 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & -3
\end{bmatrix}##

Homework Equations


##Ax=λx##

The Attempt at a Solution


I'm having some trouble finding the eigenvalues of this matrix.
The eigenvalue of a matrix is a scalar λ such that ##Ax=λx##.
To find the eigenvalue, I tried the following:

##\begin{bmatrix}
4 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & -3
\end{bmatrix} \begin{bmatrix}
x_1 \\
x_2\\
x_3
\end{bmatrix} = λ\begin{bmatrix}
x_1 \\
x_2\\
x_3
\end{bmatrix}
=
\begin{bmatrix}
4 & 0 & 0 & λ\\
0 & 0 & 0 & λ\\
1 & 0 & -3 & λ
\end{bmatrix}
##

But that doesn't seem to make any sense to me. Lambda can't be 4, 0, and -2, right?
 
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  • #2
a few things to consider

1.) you can eyeball this matrix and see this has rank 2. This means there must be an eigenvalue of 0. (why?)

2.) the trace gives the sum of the eigenvalues. ##Trace(A) = 1##, but yours sums to 2.

3.) You matrix is lower triangular. You can transpose it if you'd like as ##A## and ##A^T## have the same eigenvalues (why?). What do you know about the eigenvalues of an upper triangular matrix? Hint: think about row echelon form and pivots and singularity. Further hint: Another way to define eigenvalues are all ##\lambda## such that ##\big(A- \lambda I\big)## is singular-- This ties directly into pivots if your matrix has special structure...
 
  • #3
StoneTemplePython said:
1.) you can eyeball this matrix and see this has rank 2. This means there must be an eigenvalue of 0. (why?)

I'm afraid I don't know.

StoneTemplePython said:
2.) the trace gives the sum of the eigenvalues. Trace(A)=1Trace(A)=1Trace(A) = 1, but yours sums to 2.

I'm sorry I don't know what a trace is. We don't appear to have gotten that far yet.

StoneTemplePython said:
3.) You matrix is lower triangular.

I actually didn't realize it was triangular. So, since by definition ##x## can't be the zero vector, that means that the equation ##(A-λI)x=0## has a solution if and only if there is a free variable. ##(A-λI)## has a free variable when ##λ## is equal to ##a_{11}##, ##a_{22}##, or ##a_{33}##, meaning that the eigenvalues are 4, 0, and -3.
 
  • #4
Your view on 3 is correct. Problem solved.

Note: what this and the other problem you just posted are trying to show you is that there are other ways of exploiting special structure to more simply get at interesting eigenvalues.
- - - - -
As for point 1: consider the case where ##A## is ##n## x ##n##. if you find some ##\mathbf x \neq \mathbf 0## where ##A \mathbf x = \mathbf 0## then that means ##\mathbf x## is in the nullspace of ##A##, right?

It also means that ##A \mathbf x = 0 \mathbf x = \lambda \mathbf x = \mathbf 0##. I.e. you have found at least one case where ##\lambda = 0##. So this nullspace business is actually an eigenvalue of zero problem in disguise, for square matrices.

(It also means there is some ##\sigma = 0## -- if you have broached the subject of singular values. Feel free to disregard if you aren't familiar with singular values)
- - - - -
As for 2, trace is frequently not taught well or at all in introductory courses. It is shockingly useful when your fields is rationals, reals or complex numbers. I would strongly recommend revisiting this topic when you come across things like diagonalization, and Schur form (i.e. that every matrix over a complex field is similar to an upper triangular one... and as you just stated, we easily find the eigenvalues of an upper triangular matrix.). Trace is extensively used in quantum stuff and data science, FWIW.
 
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  • #5
Drakkith said:

Homework Statement


Find the eigenvalues of the matrix ##\begin{bmatrix}
4 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & -3
\end{bmatrix}##

Homework Equations


##Ax=λx##

The Attempt at a Solution


I'm having some trouble finding the eigenvalues of this matrix.
The eigenvalue of a matrix is a scalar λ such that ##Ax=λx##.
To find the eigenvalue, I tried the following:

##\begin{bmatrix}
4 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & -3
\end{bmatrix} \begin{bmatrix}
x_1 \\
x_2\\
x_3
\end{bmatrix} = λ\begin{bmatrix}
x_1 \\
x_2\\
x_3
\end{bmatrix}
=
\begin{bmatrix}
4 & 0 & 0 & λ\\
0 & 0 & 0 & λ\\
1 & 0 & -3 & λ
\end{bmatrix}
##

But that doesn't seem to make any sense to me. Lambda can't be 4, 0, and -2, right?
Are you familiar with the "characteristic equation" of a matrix? It is the polynomial whose roots are precisely the eigenvalues of a matrix. See, eg.,
http://golovaty.math.uakron.edu/~golovaty/fall17/linalg/common/notes/sec5_2.pdf
 
  • #6
Drakkith said:

Homework Statement


Find the eigenvalues of the matrix ##\begin{bmatrix}
4 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & -3
\end{bmatrix}##

Homework Equations


##Ax=λx##

The Attempt at a Solution


I'm having some trouble finding the eigenvalues of this matrix.
The eigenvalue of a matrix is a scalar λ such that ##Ax=λx##.
To find the eigenvalue, I tried the following:

##\begin{bmatrix}
4 & 0 & 0 \\
0 & 0 & 0 \\
1 & 0 & -3
\end{bmatrix} \begin{bmatrix}
x_1 \\
x_2\\
x_3
\end{bmatrix} = λ\begin{bmatrix}
x_1 \\
x_2\\
x_3
\end{bmatrix}
=
\begin{bmatrix}
4 & 0 & 0 & λ\\
0 & 0 & 0 & λ\\
1 & 0 & -3 & λ
\end{bmatrix}
##

But that doesn't seem to make any sense to me. Lambda can't be 4, 0, and -2, right?
Your "tableau" is incorrect; it should be
$$ \begin{array}{|ccc|c}
4 - \lambda & 0 & 0 & 0 \\
0 & -\lambda & 0 & 0 \\
1 & 0 & -3 - \lambda & 0
\end{array}
$$
 
  • #7
Drakkith said:
I'm having some trouble finding the eigenvalues of this matrix.
The eigenvalue of a matrix is a scalar λ such that Ax=λx.
To find the eigenvalue, I tried the following:
It's really, "An eigenvalue of a matrix..." not "the eigenvalue." An n x n matrix can have up to n distinct eigenvalues.

I posted an Insights article on eigenvalues and eigenvectors -- https://www.physicsforums.com/insights/what-are-eigenvectors-and-eigenvalues/.
 
  • #8
Ray Vickson said:
Are you familiar with the "characteristic equation" of a matrix?

We're just now going over that section in the book.

Mark44 said:
It's really, "An eigenvalue of a matrix..." not "the eigenvalue." An n x n matrix can have up to n distinct eigenvalues.

I knew there was (often) more than one eigenvalue for a matrix, but I hadn't realized an n x n matrix had up to n eigenvalues. Thanks Mark.
 
  • #9
Ray Vickson said:
Your "tableau" is incorrect; it should be
$$ \begin{array}{|ccc|c}
4 - \lambda & 0 & 0 & 0 \\
0 & -\lambda & 0 & 0 \\
1 & 0 & -3 - \lambda & 0
\end{array}
$$

Sorry, what's wrong with it? I swear that's the way we've done things in class.
 
  • #10
Drakkith said:
Sorry, what's wrong with it? I swear that's the way we've done things in class.

I hope that is not the way you have done it in class. Your "tableau"
$$
\begin{bmatrix}
4 & 0 & 0 & λ\\
0 & 0 & 0 & λ\\
1 & 0 & -3 & λ
\end{bmatrix}
$$
stands for the system of equations
$$ 4 x_1 = \lambda \\
0 = \lambda \\
1 x_1 - 3 x_3 = \lambda,
$$
This is not the right system.

In the tableau form, each column stands for something different: column 1 gives the coefficients of variable ##x_1##, column 2 gives coefficients for variable ##x_2##, column 3 gives the coefficients for variable ##x_3## and column 4 gives the right-hand side constants.

For the eigenvalue equations you do not yet have constants on the right, as the first right-hand-side contains the variable ##x_1##, etc. You need to first re-write the system so that all the variables are on the left and only constants are on the right, so:
$$4 x_1 - \lambda x_1 = 0 \\
0 - \lambda x_2 = 0 \\
1 x_1 - 3 x_3 - \lambda x_3 = 0
$$
In tableau form these are
$$
\begin{array}{cccc}
4 - \lambda & 0 & 0 & 0 \\
0 & -\lambda & 0 & 0 \\
1 & 0 & -3 - \lambda & 0
\end{array}
$$
It is impossible to write these with all ##\lambda##s in a single column, because the first ##\lambda## goes with ##x_1##, the second ##\lambda## goes with ##x_2##, etc. Each separate ##\lambda## needs it own, separate, column.
 
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  • #11
Ray Vickson said:
In the tableau form, each column stands for something different: column 1 gives the coefficients of variable ##x_1##, column 2 gives coefficients for variable ##x_2##, column 3 gives the coefficients for variable ##x_3## and column 4 gives the right-hand side constants.

For the eigenvalue equations you do not yet have constants on the right, as the first right-hand-side contains the variable ##x_1##, etc. You need to first re-write the system so that all the variables are on the left and only constants are on the right

You know, something like that occurred to me when I was working through the problem, but I couldn't reconcile that with the form of the matrix I had made. Now it makes sense. Thanks Ray.
 

Related to Finding the Eigenvalue of a Matrix

What is an eigenvalue?

An eigenvalue is a scalar value that represents the scale factor of a linear transformation in a given vector space.

Why is finding eigenvalues important?

Finding eigenvalues is important because it allows us to understand the behavior of a linear transformation and determine its stability and convergence.

How do you find the eigenvalues of a matrix?

The eigenvalues can be found by solving the characteristic equation of the matrix, which is det(A-λI) = 0, where A is the matrix and λ is the eigenvalue.

What is the relationship between eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are closely related, as eigenvectors are the corresponding vectors that are scaled by the eigenvalue in a linear transformation.

Can a matrix have more than one eigenvalue?

Yes, a matrix can have multiple eigenvalues. The number of eigenvalues is equal to the dimension of the matrix.

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