Question

is the matrix diagonalisable?

Answer 1

i understand how to figure out the diagonal of it but I'm not sure how to do it with a unknown number

Answer 2

@ganeshie8 could you please help? I'm really stuck on this

Answer 3

In linear algebra, a diagonal matrix is a matrix (usually a square matrix) in which the entries outside the main diagonal (↘) are all zero. The diagonal entries themselves may or may not be zero. Thus, the matrix D = (di,j) with n columns and n rows is diagonal if:

d_{i,j} = 0 \text{ if } i \ne j\ \forall i,j \in \{1, 2, \ldots, n\}
For example, the following matrix is diagonal:

\begin{bmatrix}
1 & 0 & 0\\
0 & 4 & 0\\
0 & 0 & -2\end{bmatrix}
The term diagonal matrix may sometimes refer to a rectangular diagonal matrix, which is an m-by-n matrix with all the entries not of the form di,i being zero. For example:

\begin{bmatrix}
1 & 0 & 0\\
0 & 4 & 0\\
0 & 0 & -3\\
0 & 0 & 0\\
\end{bmatrix} or \begin{bmatrix}
1 & 0 & 0 & 0 & 0\\
0 & 4 & 0& 0 & 0\\
0 & 0 & -3& 0 & 0\end{bmatrix}

Answer 4

Hints:
1. A matrix is diagonalizable when it has distinct eigenvalues.
2. A triangular matrix has its eigenvalues on the main diagonal.

Answer 5

@mathmate
what is eigenvalues mean

Answer 6

@dayakar a root of the characteristic equation

Answer 7

@swift_13 do you need the solution/?

Answer 8

We can say that if the unknown number is 0 for both cases then the matrices are diagonalisable since they are already diagonal already. For other cases I have to work it out.

Answer 9

Try writing down the matrices \(\lambda I-A\) and see what you get.

Answer 10

@dayakar
Eigenvalues are the solution to the characteristic equation of the given matrix A, \(det (\lambda I-A)=0\) where det() means the "determinant of".
For more information, read Paul's tutorial on the subject, or google other articles:
http://tutorial.math.lamar.edu/Classes/DE/LA_Eigen.aspx