MHB Diagonalizable Matrix: How to Approach?

[Yankel]

Hello all

I have this matrix:

\[\begin{pmatrix} 6 & 0\\ -3 & a \end{pmatrix}\]

And I am told it is diagonalizable. Therefore, the value of a is:

1) a=0
2) a not= 0
3) a not=6
4) a=6
5) a not=0,6

How should I approach this? Is there a "trick" or should I find eigenvalues and eigenvectors for both values of a?

Thank you.

 

[I like Serena]

Hello all

I have this matrix:

\[\begin{pmatrix} 6 & 0\\ -3 & a \end{pmatrix}\]

And I am told it is diagonalizable. Therefore, the value of a is:

1) a=0
2) a not= 0
3) a not=6
4) a=6
5) a not=0,6

How should I approach this? Is there a "trick" or should I find eigenvalues and eigenvectors for both values of a?

Thank you.

Hi Yankel! ;)

Are you familiar with the Jordan normal form?
If so, what are the possibilities for the Jordan normal form?

Furthermore, a "trick" is that the trace ($6+a$) is equal to the sum of the eigenvalues, and the determinant is equal to the product of the eigenvalues.

 

[Yankel]

Hi,

So 6a is the product of the eigenvalues and 6+a is the sum.

if a=0, the product is 0 and the sum is 6

if a=6 the product is 36 and the sum is 12

how does that help me find a ?

what is Jordan normal form?

 

[I like Serena]

Hi,

So 6a is the product of the eigenvalues and 6+a is the sum.

if a=0, the product is 0 and the sum is 6

if a=6 the product is 36 and the sum is 12

how does that help me find a ?

what is Jordan normal form?

More specifically, the eigenvalues are $6$ and $a$.

And if you're not familiar with the Jordan normal form, you're probably supposed to indeed find the eigenvalues and the eigenvectors to do the diagonalization.

For the record, every matrix is similar to a Jordan normal form.
It tells us something about how diagonalizable a matrix is, and how to categorize matrices..
Note that a diagonalizable matrix is defined as a matrix that is similar to a diagonal matrix.

The possibilities for the Jordan normal form are:
$$\begin{pmatrix}\lambda_1 & 1 \\ 0 & \lambda_1\end{pmatrix},
\begin{pmatrix}\lambda_1 & 0 \\ 0 & \lambda_1\end{pmatrix},
\begin{pmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{pmatrix}
$$
It means that the matrix is guaranteed to be diagonalizable if the eigenvalues are different.
And if the eigenvalues are the same, we cannot tell yet. We'll have to check further.

 

[HallsofIvy]

A matrix is "diagonalizable" if and only if it has a "complete set" of eigenvectors. That is, that there exist a basis for the vector space consisting of Eigen vectors. Here, that means that there must be two independent eigenvectors.

It is obvious that the two eigenvalues are "6" and "a". If a is any number other than "6" the two eigenvectors will be independent (eigenvectors corresponding to distinct eigenvalues are always independent). Since this a "triangular" matrix, it is obvious that if a= 6, there are not two independent eignvectors.