Question

Find all (real) values of k for which A is diagonalizable. (Enter your answers as a comma-separated list.)A = [4 k0 4 ] k = ___.

Answer 1

Answer:

We have the matrix

$A=\left[\begin{array}{cc}4&k\\0&4\end{array}\right]$

Remember, every symmetric matrix with real coefficients is diagonalizable. Note that the coefficients of A are reals. So, if we take $k$ such that A is symmetric, then A is diagonalizable.

For A to be symmetric, $k = a_ {12} = a_ {21} = 0$

Then $k = 0$.

Answer 2

Answer: K needs to be equal to zero.

Step-by-step explanation:

The matrix is:

$A = \left[\begin{array}{ccc}4&k\\0&4\end{array}\right]$

A matrix is diagonalizable if

A = P*D*P^(-1)

Where D is a diagonalized matrix formed with the eigenvalues of A, and P is a matrix formed with the eigenvectors of A.

We want to find the values of k that allow us to diagonalize this matrix.

A matrix is diagonalizable if:

The eigenvalues of A can be calculated by:

P(a) = det(A - x*I) = (4 - x)^2 =  the eigenvalues are 4 two times.

$D = \left[\begin{array}{ccc}4&0\\0&4\end{array}\right]$

then eigenvectors are such:

A*v = 4*v

So the eigenvectors are:

$v1 = \left[\begin{array}{ccc}1\\0\end{array}\right] , v2 = \left[\begin{array}{ccc}0\\1\end{array}\right] ,$

then we will have that:

$P = v1,v2 = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right] = I$

and, P^-1 = I^-1 = I

then:

A = I*D*I

every matrix multiplied by the identity is the same matrix:

A = I*D*I = D

then we have that A must be equal to D, then k must be zero,