Matrix with repeated eigenvalues

[Shambles]

The question is asking for what values of x will the matrix have at least one repeated eigenvalue (algebraic multiplicity of 2 or greater). The matrix is
| 3 0 0 |
| 0 x 2 | So naturally a normal attempt to find the eigenvalue in a question with only intergers
| 0 2 x | I would continue with:

| $$\lambda$$-3 0 0 |
| 0 $$\lambda$$-x -2|
| 0 -2 $$\lambda$$-x|

And then finding the determinant would continue with ($$\lambda$$-3)($$\lambda$$-x)($$\lambda$$-x) - (-2)(-2)($$\lambda$$-3) etc...

Except with 2 unknown variable it inevitably becomes a problem that I run into with more theoretical questions vs questions dealing only with numbers. As far as I know an eigenvalue with an algebraic multiplicity of >1 doesn't even have a geometric significance. Clearly I am approaching this entire question from the wrong angle and could use a push in the right direction if anyone is so able. Muchly appreciated.

 

[HallsofIvy]

I don't see why you need to be that theoretical!

The characteristic equation for the eigenvalues is obviously [itex](\lambda- 3)((\lambda- x)^2- 4)= 0. 3 is an obvious solution to that so in order to have a double root, either 3 is a root of $(\lambda- x)^2- 4= 0$ or that equation has a double root. If 3 is a root then we must have $(3- x)^2- 4= x^2- 6x+ 5= (x- 3)(x- 2)= 0$ so that x= 2 or 3. If $(\lambda- x)^2- 4= \lambda^2- 2x\lambda+ x^2- 4= 0$ has a double root, then its discriminant must be 0: $4x^2- 4(x^2- 4)= 4$ is never 0 so that can't happen.

 

[Architeuthis Dux]

I would approach this problem by first understanding the concept of eigenvalues and their multiplicity. An eigenvalue represents a scalar value that, when multiplied with a vector, gives a new vector in the same direction. The algebraic multiplicity of an eigenvalue represents the number of times it appears as a root of the characteristic polynomial of the matrix.

In this matrix, we have two variables, x and λ. To find the values of x that will result in at least one repeated eigenvalue, we need to solve the characteristic polynomial. However, since we have two unknown variables, we cannot find a specific value for x that will result in a repeated eigenvalue. Instead, we can find a range of values for x that will result in at least one repeated eigenvalue.

To do this, we can use the fact that the sum of the eigenvalues of a matrix is equal to the trace of the matrix. In this case, the trace is 3 + x + 2 = x + 5. So, the sum of the eigenvalues will be equal to x + 5. If we assume that x is a real number, then the sum of the eigenvalues must also be a real number.

Now, we know that the sum of the eigenvalues must be equal to x + 5 and we want at least one repeated eigenvalue. This means that one of the eigenvalues must be equal to x + 5 and the other two eigenvalues must be equal to some other number. In other words, we need to find a range of values for x that will result in at least one repeated eigenvalue.

To do this, we can use the fact that the product of the eigenvalues is equal to the determinant of the matrix. In this case, the determinant is (3-x)(x^2-4). So, the product of the eigenvalues will be equal to (3-x)(x^2-4). If we assume that x is a real number, then the product of the eigenvalues must also be a real number.

Now, we know that the product of the eigenvalues must be equal to (3-x)(x^2-4) and we want at least one repeated eigenvalue. This means that one of the eigenvalues must be equal to (3-x)(x^2-4) and the other two eigenvalues must be equal to some other number. In other words, we need to find