Eigenvalues of matrix containing an unknown

3,199

Solution 1

To confirm that $\lambda$ is an eigenvalue of a matrix $M$ you do not need to determine its characteristic equation; all you have to do is to write down an eigenvector, that is, find a $v$ such that $Mv=\lambda v$, equivalently $(M-\lambda I)v=0$.

In your case, if $\lambda=a+2$ then $$M-\lambda I=\pmatrix{-2&1&1\\1&-2&1\\1&1&-2}.$$ Now can you find a nonzero vector with $(M-\lambda I)v=0$? (Ah, $M-\lambda I$ doesn't have an $a$ in it!)

Solution 2

It’s often easier for matrices that show obvious patterns to start by finding eigenvectors.

Observe that the row sums of this matrix (call it $A$) are all equal to $a+2$. Summing across rows is equivalent to multiplying by $(1,1,1)^T$, so we know that this is an eigenvector of the matrix with eigenvalue $a+2$. That’s one down.

For the other eigenvalue, notice that $A-(a-1)I$ consists of all ones, and so is singular, but that’s exactly the condition for $a-1$ to be an eigenvalue. The null space of $A-(a-1)I$ is clearly two-dimensional, so $a-1$ has geometric multiplicity $2$ and we’ve accounted for all of the eigenvalues. Note that if we’d done this first, we could instead have found the remaining eigenvalue $a+2$ by using the fact that the trace of the matrix is equal to the sum of its eigenvalues: $$\lambda+2(a-1)=3a\implies\lambda=a+2.$$

Solution 3

Here's another way to quickly find the eigenvalues of that matrix: Obviously, $$\pmatrix{a&1&1\\1&a&1\\1&1&a} = \pmatrix{1&1&1\\1&1&1\\1&1&1} + (a-1)I=:C+(a-1)I$$ where $I$ is the identity matrix. Note that adding a multiple of the identity matrix doesn't change the eigenvectors, and changes the eigenvalues by the prefactor, in this case $(a-1)$.

Clearly, $C$ has rank $1$ (just subtract the first row from all others), so all but one eigenvalue is zero, and thus all but one eigenvalue of the original matrix is $a-1$. Also, it is quite obvious that $\pmatrix{1&1&1}^T$ is eigenvector of $C$ to the eigenvalue $3$, thus it is also eigenvector of the original matrix to the eigenvalue $3+(a-1)=a+2$.

Since the matrix is real symmetric, the eigenspaces are orthogonal, and thus the eigenspace for the eigenvalue $a-1$ consists of all vectors that are orthogonal to $\pmatrix{1&1&1}^T$, that is, of all vectors whose components add to $0$.

Solution 4

Let $A = \begin{bmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 &a \end{bmatrix} $

To find the eigenvalues, you solve :

$\det(A-Iλ) = 0 \Leftrightarrow \begin{vmatrix} a-λ & 1 & 1 \\ 1 & a-λ & 1 \\ 1 & 1 &a-λ \end{vmatrix}=0 \Leftrightarrow $

$ \Leftrightarrow a^3 - 3 a^2 λ + 3 a λ^2 - 3 a -λ^3 + 3 λ + 2 = 0 \Leftrightarrow$

$\Leftrightarrow 2 - 3 a + a^3 + λ (3 - 3 λ^2) + 3 a λ^2 - λ^3 = 0\Leftrightarrow $

$\Leftrightarrow (a - λ - 1)^2 (a - λ + 2) = 0 \Leftrightarrow $

$\Leftrightarrow λ=a-1 $ or $ λ=a+2$

Note that the eigenvalue : $λ=a-1$ is of double multiplicity.

I factorized the equation after collecting in terms of $λ$ and observing the factorization of the product.

Share:
3,199

Related videos on Youtube

Olivier Perrault
Author by

Olivier Perrault

Updated on April 20, 2020

Comments

  • Olivier Perrault
    Olivier Perrault over 3 years

    hey I am having a hard time finding the eigenvalues of such a matrix.
    My homework asks me to show that the eigenvalues are $\lambda_1 = a+2$, $\lambda_2 = a-1$;

    \begin{bmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \end{bmatrix}

    I have manage determine that the determinant is $(a-\lambda)^3 - (a-\lambda) +2$;

    However I fail to reduce it further to be able to factor it and obtain the eigenvalues.
    Can you guys help me.

    • celtschk
      celtschk over 6 years
      Just insert $\lambda_1$ and $\lambda_2$ into the determinant and verify that you get $0$. Then use polynomial division to see that no different third eigenvalue exists.
  • amd
    amd over 6 years
    While you can certainly verify that $a+2$ and $a-1$ are eigenvalues of the matrix this way, how do you know that those are the only ones? It’s a $3\times3$ matrix, so there might be a third eigenvalue not equal to these two.
  • Angina Seng
    Angina Seng over 6 years
    @amd When you have two, the third comes for free: the trace of a matrix is the sum of its eigenvalues.
  • amd
    amd over 6 years
    You might add that to your answer to make it complete. The problem, after all, wasn’t only to verify the two given eigenvalues.