Find the eigenvalues and eigenvectors

[Mutatis]

Homework Statement

Find the eigenvalues and eigenvectors of the following matrix: $$ A = \begin{bmatrix}
3 & 0 & 0 \\
0 & 3 & 2 \\
0 & -1 & 0
\end{bmatrix}
$$

Homework Equations

Characteristic polynomial: $$ \Delta (t) = t^3 - Tr(A) t^2 + (A_{11}+A_{22} +A_{33})t - det(A) .$$

The Attempt at a Solution

From the characteristic polynomial equation above, I've found the characteristic polynomial of the matrix and then I'd found the eigenvalues: $$ \lambda_1 = 2 \\ \lambda_2 = \frac {7 + \sqrt {61}} {2} \\ \lambda_3 = \frac {7 - \sqrt {61}} {2} .$$
But these values of $\lambda$ are too strange though. I've tried to calculate this using Mathway calculator, but there aren't no answer for this problem there. I think this problem hasn't a solution. What do you guys think about it?

 

[fresh_42]

The characteristic polynomial cannot be correct, as it has the coefficient trace twice. Use the definition as $\Delta (t)=\det \left( A-t\cdot I \right)$ instead. Obviously, $3$ is an eigenvalue, and the determinant can be calculated blockwise.

 

[Ray Vickson]

Homework Statement

Find the eigenvalues and eigenvectors of the following matrix: $$ A = \begin{bmatrix}
3 & 0 & 0 \\
0 & 3 & 2 \\
0 & -1 & 0
\end{bmatrix}
$$

Homework Equations

Characteristic polynomial: $$ \Delta (t) = t^3 - Tr(A) t^2 + (A_{11}+A_{22} +A_{33})t - det(A) .$$

The Attempt at a Solution

From the characteristic polynomial equation above, I've found the characteristic polynomial of the matrix and then I'd found the eigenvalues: $$ \lambda_1 = 2 \\ \lambda_2 = \frac {7 + \sqrt {61}} {2} \\ \lambda_3 = \frac {7 - \sqrt {61}} {2} .$$
But these values of $\lambda$ are too strange though. I've tried to calculate this using Mathway calculator, but there aren't no answer for this problem there. I think this problem hasn't a solution. What do you guys think about it?

I would not call these values strange, but I would call them wrong. Write out the actual numerical coefficients in your characteristic polynomial, because something is amiss in your work.

 

[Mark44]

I agree with fresh_42 and Ray that there are errors in your work. One of your eigenvalues is correct -- $\lambda_1 = 2$ and the others are not. All three eigenvalues are integers.and all are distinct.

Thread moved -- please post questions about eigenvalues and eigenvectors in the Calculus & Beyond section. These topics are usually taught after calculus.

 

[Mutatis]

The characteristic polynomial cannot be correct, as it has the coefficient trace twice. Use the definition as $\Delta (t)=\det \left( A-t\cdot I \right)$ instead. Obviously, $3$ is an eigenvalue, and the determinant can be calculated blockwise.

Yes, I did my calculation wrong. I'd computed $3+3=9$ instead of $3+3=6$. Now I got it right, my eigenvalues are $\lambda_1 = 2, \lambda_2 = -1$ and $\lambda_3 = -3$.

I agree with fresh_42 and Ray that there are errors in your work. One of your eigenvalues is correct -- $\lambda_1 = 2$ and the others are not. All three eigenvalues are integers.and all are distinct.

Thread moved -- please post questions about eigenvalues and eigenvectors in the Calculus & Beyond section. These topics are usually taught after calculus.

Sorry for that, I'll post it at the right place next time. Thank all of you, by the way.

 

[Mark44]

Yes, I did my calculation wrong. I'd computed $3+3=9$ instead of $3+3=6$. Now I got it right, my eigenvalues are $\lambda_1 = 2, \lambda_2 = -1$ and $\lambda_3 = -3$..

I get different values for $\lambda_2$ and $\lambda_3$.
I've gone ahead and calculated the eigenvectors for all three of my eigenvalues, and they check out, so I'm confident of my results.

Because of the many arithmetic calculations that are involved in these kinds of problems, it's always a good idea to verify that your answers are correct. If $\lambda$ is an eigenvalue of matrix A, and $\vec x$ is the associated eigenvalue, then $A\vec x = \lambda \vec x$ should be a true statement.

 

[StoneTemplePython]

Yes, I did my calculation wrong. I'd computed $3+3=9$ instead of $3+3=6$. Now I got it right, my eigenvalues are $\lambda_1 = 2, \lambda_2 = -1$ and $\lambda_3 = -3$.

something else you should know: you have real scalars and matrix $A$
$A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 2 \\ 0 & -1 & 0 \end{bmatrix}$

and you even mentioned the trace in your original post (though there were issues with how it was used).

A very easy sanity check on your work is: compute the trace and know that it gives the sum of the eigenvalues (why?), so

$6 = 3 + 3 +0 = \text{trace}\big(A\big) = \lambda_1 + \lambda_2 + \lambda_3 \neq \lambda_1 + \lambda_2 + \lambda_3 = 2 + -1 + -3 = -2$

so there must be something wrong here.
- - - - -
note that computing the trace is very easy and it would have redflagged your original answer as well, because

$6 \neq 2 + \frac {7 + \sqrt {61}} {2} + \frac {7 - \sqrt {61}} {2} = 2 + 7 = 9$

 

[Mutatis]

I get different values for $\lambda_2$ and $\lambda_3$.
I've gone ahead and calculated the eigenvectors for all three of my eigenvalues, and they check out, so I'm confident of my results.

Because of the many arithmetic calculations that are involved in these kinds of problems, it's always a good idea to verify that your answers are correct. If $\lambda$ is an eigenvalue of matrix A, and $\vec x$ is the associated eigenvalue, then $A\vec x = \lambda \vec x$ should be a true statement.

Oh my God... I've done wrong again. The right answer for the eigenvalues is $\lambda_1= 2, \lambda_2 = 1$ and $\lambda_3 = 3$!

something else you should know: you have real scalars and matrix $A$
$A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 2 \\ 0 & -1 & 0 \end{bmatrix}$

and you even mentioned the trace in your original post (though there were issues with how it was used).

A very easy sanity check on your work is: compute the trace and know that it gives the sum of the eigenvalues (why?), so

$6 = 3 + 3 +0 = \text{trace}\big(A\big) = \lambda_1 + \lambda_2 + \lambda_3 \neq \lambda_1 + \lambda_2 + \lambda_3 = 2 + -1 + -3 = -2$

so there must be something wrong here.
- - - - -
note that computing the trace is very easy and it would have redflagged your original answer as well, because

$6 \neq 2 + \frac {7 + \sqrt {61}} {2} + \frac {7 - \sqrt {61}} {2} = 2 + 7 = 9$

Thank you! I'm going to check my calculations before freaking out. I'm so impulsive...

 

[StoneTemplePython]

Thank you! I'm going to check my calculations before freaking out. I'm so impulsive...

I like easy stuff. To my mind, there's one major loose end here:

Obviously, $3$ is an eigenvalue

Why is this true/obvious?

Put differently, I can eyeball your matrix and it just jumps out at me, which is a lot more pleasant than getting deep in the weeds in some characteristic polynomial or determinant calculation. Remember: being an eigenvalue $\lambda$ is that same thing as saying the matrix $\big(\lambda I - A\big) = \begin{bmatrix} \lambda - 3 & 0 & 0 \\ 0 & \lambda - 3 & -2 \\ 0 & 1 & \lambda - 0 \end{bmatrix}$

is singular / not invertible.

 

[fresh_42]

Another simple way to see it is, that $A=\left[\begin{array}{c|cc} 3 & 0 & 0 \\ \hline 0 & -3 & -2 \\ 0 & 1 & 0 \\ \end{array}\right]$ and with $B=\begin{bmatrix}-3&-2\\1&0\end{bmatrix}$ we have $A= \left[\begin{array}{c|ccc} 3 & 0 && 0 \\ \hline 0 & \\ &&B&\\ 0 &&& \\ \end{array}\right]$ and it can directly be seen that $A.(1,0,0)^\tau = 3\cdot (1,0,0)^\tau$ and $(1,0,0)$ is an eigenvector to the eigenvalue $3$.
Additionally we have $\det(A-\lambda \cdot I_3) = (3-\lambda) \cdot \det (B-\lambda I_2)\,.$ Then $\det (B-\lambda I_2)$ is just a quadratic polynomial.

 

[Ray Vickson]

I like easy stuff. To my mind, there's one major loose end here:Why is this true/obvious?

Put differently, I can eyeball your matrix and it just jumps out at me, which is a lot more pleasant than getting deep in the weeds in some characteristic polynomial or determinant calculation. Remember: being an eigenvalue $\lambda$ is that same thing as saying the matrix $\big(\lambda I - A\big) = \begin{bmatrix} \lambda - 3 & 0 & 0 \\ 0 & \lambda - 3 & -2 \\ 0 & 1 & \lambda - 0 \end{bmatrix}$

is singular / not invertible.

It is also the same as saying that the determinant $\det(\lambda I - A) = 0,$ and because the first row/column consists of zero in the off-diagonal, we have
$$\det = (\lambda - 3) \times \text{something}.$$ Here, the "something" is the $2 \times 2$ southeast determinant.

 

[StoneTemplePython]

It is also the same as saying that the determinant $\det(\lambda I - A) = 0,$ and because the first row/column consists of zero in the off-diagonal, we have
$$\det = (\lambda - 3) \times \text{something}.$$ Here, the "something" is the $2 \times 2$ southeast determinant.

Yes, I agree of course. I happen to like determinants, but for me and eyeballing things, the most immediate thing is to see that selecting $\lambda := 3$ means the above square matrix has a column (and even a row) of all zeros -- such a matrix can never be full rank / must have linearly dependent columns/ is singular, etc. I mean I just see the zero vector sitting there...

 

[Mutatis]

Hey guys, I'm here again... I don't know why I'm still having troubles with this kind of subject. I did my readings and so on, but I'm still struggling to get it right... Look, I got another exercise here, I need to find the eigenvalues and eigenvectors of: $$ \begin{bmatrix}
4 & -2 & 0 \\
-1 & 1 & 0 \\
0 & 1 & 2
\end{bmatrix}. $$
To find the eigenvalues I've done what you'd said above, $det \left( A - \lambda I \right) = 0$ to get the characteristic polinomial $$\Delta (t) = t^3 - 7t^2 + 12t-4.$$
But I've only found out one of the eigenvalues through this equation, $\lambda_1 = 2$. The others values I couldn't find because they aren't integers.

 

[vela]

The other two eigenvalues are the roots of a quadratic polynomial, so use the quadratic equation.

 

[Mutatis]

The other two eigenvalues are the roots of a quadratic polynomial, so use the quadratic equation.

Yes, $\left( t-2\right) \left(t^2-5t+2\right)$, with $\lambda_2 = \frac {5} {2} - \frac {\sqrt {17}} {2}$ and $\lambda_3 = \frac {5} {2} + \frac {\sqrt {17}} {2}$.

 

[Mark44]

Yes, $\left( t-2\right) \left(t^2-5t+2\right)$, with $\lambda_2 = \frac {5} {2} - \frac {\sqrt {17}} {2}$ and $\lambda_3 = \frac {5} {2} + \frac {\sqrt {17}} {2}$.

These look fine. Since the eigenvalues $\lambda_2$ and $\lambda_3$ are complex conjugates, so will be their associated eigenvectors. It's a bit tedious and error-prone to find the eigenvector associated with either of these eigenvalues, at least you don't have to do the same work twice.

 

[Mutatis]

These look fine. Since the eigenvalues $\lambda_2$ and $\lambda_3$ are complex conjugates, so will be their associated eigenvectors. It's a bit tedious and error-prone to find the eigenvector associated with either of these eigenvalues, at least you don't have to do the same work twice.

The eigenvector associated to these eigenvalues are $\vec v_1 = (0,0,0) , \vec v_2 = (0,0,0)$... That's what I've found out.

 

[fresh_42]

The eigenvector associated to these eigenvalues are $\vec v_1 = (0,0,0) , \vec v_2 = (0,0,0)$... That's what I've found out.

The zero vector is always in the eigenspace to an eigenvalue, but it is usually not called an eigenvector. The latter is assumed to be non trivial. Have you solved the equation $$\begin{bmatrix}4&-2&0\\-1&1&0\\0&1&2\end{bmatrix} \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}\lambda x\\ \lambda y\\ \lambda z\end{bmatrix}$$
with the given values for the three eigenvalues $\lambda\,?$

 

[Mutatis]

The zero vector is always in the eigenspace to an eigenvalue, but it is usually not called an eigenvector. The latter is assumed to be non trivial. Have you solved the equation $$\begin{bmatrix}4&-2&0\\-1&1&0\\0&1&2\end{bmatrix} \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}\lambda x\\ \lambda y\\ \lambda z\end{bmatrix}$$
with the given values for the three eigenvalues $\lambda\,?$

Yes, the values are $\left( t-2\right) \left(t^2-5t+2\right)$, with $\lambda_1 = 2, \lambda_2 = \frac {5} {2} - \frac {\sqrt {17}} {2}$ and $\lambda_3 = \frac {5} {2} + \frac {\sqrt {17}} {2}$.

 

[Ray Vickson]

Yes, the values are $\left( t-2\right) \left(t^2-5t+2\right)$, with $\lambda_1 = 2, \lambda_2 = \frac {5} {2} - \frac {\sqrt {17}} {2}$ and $\lambda_3 = \frac {5} {2} + \frac {\sqrt {17}} {2}$.

If you are replying to post #18, then you are answering the wrong question. You were asked what the values of $x,y,z$ are, not what is the value of $\lambda.$

To keep threads manageable, please employ the "reply" button, so we can tell which response goes with which message---otherwise we soon have chaos.