MHB Find Eigenvalues of 4x4 Matrix with -1 & 3 Entries

[A.Magnus]

Could anyone help me figure out the eigenvalues of a $4 \times 4$ matrix, with all entries being $-1$ except the entries of non-main diagonal being $3$?

$$ A =
\begin{pmatrix}
3 &-1 &-1 &-1\\
-1 &3 &-1 &-1\\
-1 &-1 &3 &-1\\
-1 &-1 &-1 &3\\
\end{pmatrix}$$

For small matrices, I would use characteristic equation $|A - \lambda I| = 0$ without hesitation, but this matrix is too big. Thank you before hand for your gracious helping hand and time. ~MA

 

[Prove It]

Could anyone help me figure out the eigenvalues of a $4 \times 4$ matrix, with all entries being $-1$ except the entries of non-main diagonal being $3$?

$$ A =
\begin{pmatrix}
3 &-1 &-1 &-1\\
-1 &3 &-1 &-1\\
-1 &-1 &3 &-1\\
-1 &-1 &-1 &3\\
\end{pmatrix}$$

For small matrices, I would use characteristic equation $|A - \lambda I| = 0$ without hesitation, but this matrix is too big. Thank you before hand for your gracious helping hand and time. ~MA

The matrix you have written has 3s on the MAIN diagonal...

If you don't want to use the characteristic equation (which will still work) you could diagonalise the matrix. Then use the fact that the determinant of a diagonal matrix is equal to the product of the terms on the main diagonal...

 

[I like Serena]

Hey MaryAnn! (Wave)

Well, with a computer we can do singular value decomposition, but I believe that by hand the characteristic equation is as easy as it gets.
And we have Wolfram|Alpha to help us.

We do have the properties that the trace (sum of the elements on the diagonal) is equal to the sum of the eigenvalues.
And the determinant is equal to the product of the eigenvalues.
So we can use that to verify we didn't make a mistake.

Actually, we can use those properties to find the eigenvalues of $$\begin{bmatrix}3&-1\\-1&3\end{bmatrix}$$
So we have that $t_1+t_2=3+3=6$ and $t_1 t_2 = 3\cdot 3- -1 \cdot -1= 8$.
Now we can formulate it as the quadratic equation $t^2 - 6t +8=0$.
That's because the sum of the roots of $x^2+bx+c=0$ is $-b$ and the product of the roots is $c$.
Solving it gives us $t_1=2,t_2=4$. (Cool)

 

[A.Magnus]

The matrix you have written has 3s on the MAIN diagonal...

If you don't want to use the characteristic equation (which will still work) you could diagonalise the matrix. Then use the fact that the determinant of a diagonal matrix is equal to the product of the terms on the main diagonal...

I am sorry for the mixed up! It was indeed the MAIN diagonal. Your idea about diagonalizing the matrix is very interesting, but my textbook says that to diagonalize a matrix, first you need to find out its eigenvalue first? Am I getting mixed up again here? Am not getting correct again? Let me know and thanks for your gracious help and time, as always. ~MA

 

[A.Magnus]

Hey MaryAnn! (Wave)

Well, with a computer we can do singular value decomposition, but I believe that by hand the characteristic equation is as easy as it gets.
And we have Wolfram|Alpha to help us.

We do have the properties that the trace (sum of the elements on the diagonal) is equal to the sum of the eigenvalues.
And the determinant is equal to the product of the eigenvalues.
So we can use that to verify we didn't make a mistake.

Actually, we can use those properties to find the eigenvalues of $$\begin{bmatrix}3&-1\\-1&3\end{bmatrix}$$
So we have that $t_1+t_2=3+3=6$ and $t_1 t_2 = 3\cdot 3- -1 \cdot -1= 8$.
Now we can formulate it as the quadratic equation $t^2 - 6t +8=0$.
That's because the sum of the roots of $x^2+bx+c=0$ is $-b$ and the product of the roots is $c$.
Solving it gives us $t_1=2,t_2=4$. (Cool)

As always, many thanks to your hints. I saw from your link that the characteristic polynomial is

$$\lambda^4 - 12 \lambda^3 + 48 \lambda^2 - 64 \lambda,$$

but how do you break it down to its determinant and trace? In other words, what is the general characteristic polynomial for a $4 \times 4$ matrix? I have been looking up for help from this Wikipedia site, but I am still getting confused. Any more help? Thank again for your gracious help and time, as always. ~MA

 

[Opalg]

I think that the simplest way to do this is to use elementary row (or column) operations on the determinant. You want to calculate $$|A-\lambda I| = \begin{vmatrix}3 - \lambda &-1 &-1 &-1\\ -1 &3 - \lambda &-1 &-1\\ -1 &-1 &3 - \lambda &-1\\ -1 &-1 &-1 &3 - \lambda\\ \end{vmatrix}.$$ Add all the other rows to the top row, and this becomes $$|A-\lambda I| = \begin{vmatrix}- \lambda &- \lambda &- \lambda &- \lambda \\ -1 &3 - \lambda &-1 &-1\\ -1 &-1 &3 - \lambda &-1\\ -1 &-1 &-1 &3 - \lambda\\ \end{vmatrix}.$$ Now you can take a factor $-\lambda$ out of the top row, to get $$|A-\lambda I| = -\lambda\begin{vmatrix}1&1&1&1\\ -1 &3 - \lambda &-1 &-1\\ -1 &-1 &3 - \lambda &-1\\ -1 &-1 &-1 &3 - \lambda\\ \end{vmatrix}.$$ The next step will be to subtract the left column from each of the other columns. The top row will then become $1\ 0\ 0\ 0$, so you can easily expand the determinant along the top row. Continue in that way, and you will get the characteristic equation (in an already factorised form).

 

[A.Magnus]

I think that the simplest way to do this is to use elementary row (or column) operations on the determinant. You want to calculate $$|A-\lambda I| = \begin{vmatrix}3 - \lambda &-1 &-1 &-1\\ -1 &3 - \lambda &-1 &-1\\ -1 &-1 &3 - \lambda &-1\\ -1 &-1 &-1 &3 - \lambda\\ \end{vmatrix}.$$ Add all the other rows to the top row, and this becomes $$|A-\lambda I| = \begin{vmatrix}- \lambda &- \lambda &- \lambda &- \lambda \\ -1 &3 - \lambda &-1 &-1\\ -1 &-1 &3 - \lambda &-1\\ -1 &-1 &-1 &3 - \lambda\\ \end{vmatrix}.$$ Now you can take a factor $-\lambda$ out of the top row, to get $$|A-\lambda I| = -\lambda\begin{vmatrix}1&1&1&1\\ -1 &3 - \lambda &-1 &-1\\ -1 &-1 &3 - \lambda &-1\\ -1 &-1 &-1 &3 - \lambda\\ \end{vmatrix}.$$ The next step will be to subtract the left column from each of the other columns. The top row will then become $1\ 0\ 0\ 0$, so you can easily expand the determinant along the top row. Continue in that way, and you will get the characteristic equation (in an already factorised form).

I think this solution is the one, the best one! Wish I could thank you twice! Again and again, thanks for all your graciousness! ~MA

 

[Prove It]

I am sorry for the mixed up! It was indeed the MAIN diagonal. Your idea about diagonalizing the matrix is very interesting, but my textbook says that to diagonalize a matrix, first you need to find out its eigenvalue first? Am I getting mixed up again here? Am not getting correct again? Let me know and thanks for your gracious help and time, as always. ~MA

I meant using row operations on the matrix itself until it is upper triangularised...

 

[A.Magnus]

I meant using row operations on the matrix itself until it is upper triangularised...

Thank you. ~MA