Questions: TRUE or FALSE: The matrix A=[2 1; 6 3] is diagonalizable TRUE because the vectors [1; 3] and [2; -1] form an eigenbasis for A FALSE because det(A)=0 FALSE because A has rank 1 TRUE because the vectors and [-1; 2] and [1; 3] form an eigenbasis for A

TRUE or FALSE: The matrix A=[2 1; 6 3] is diagonalizable
TRUE because the vectors [1; 3] and [2; -1] form an eigenbasis for A
FALSE because det(A)=0
FALSE because A has rank 1
TRUE because the vectors and [-1; 2] and [1; 3] form an eigenbasis for A

Transcript text: TRUE or FALSE: The matrix $A=\left[\begin{array}{ll}2 & 1 \\ 6 & 3\end{array}\right]$ is diagonalizable TRUE because the vectors $\left[\begin{array}{l}1 \\ 3\end{array}\right]$ and $\left[\begin{array}{c}2 \\ -1\end{array}\right]$ form an eigenbasis for $A$ FALSE because $\operatorname{det}(A)=0$ FALSE because $A$ has rank 1 TRUE because the vectors and $\left[\begin{array}{c}-1 \\ 2\end{array}\right]$ and $\left[\begin{array}{l}1 \\ 3\end{array}\right]$ form an eigenbasis for $A"

Solution

Solution Steps

To determine if the matrix $ A = \begin{bmatrix} 2 & 1 \\ 6 & 3 \end{bmatrix} $ is diagonalizable, we need to check if it has a full set of linearly independent eigenvectors. This involves finding the eigenvalues and corresponding eigenvectors of $ A $. If the number of linearly independent eigenvectors equals the size of the matrix (which is 2 in this case), then $ A $ is diagonalizable.

Step 1: Determine Eigenvalues

The eigenvalues of the matrix $ A = \begin{bmatrix} 2 & 1 \\ 6 & 3 \end{bmatrix} $ are calculated to be $ \lambda_1 = 0 $ and $ \lambda_2 = 5 $.

Step 2: Determine Eigenvectors

The corresponding eigenvectors for the eigenvalues are:

For $ \lambda_1 = 0 $: The eigenvector is approximately $ \begin{bmatrix} -0.4472 \\ 0.8944 \end{bmatrix} $.
For $ \lambda_2 = 5 $: The eigenvector is approximately $ \begin{bmatrix} -0.3162 \\ -0.9487 \end{bmatrix} $.

Step 3: Check Linear Independence

The rank of the matrix formed by the eigenvectors is $ 2 $, which is equal to the size of the matrix $ A $ (which is $ 2 \times 2 $). This indicates that the eigenvectors are linearly independent.

Step 4: Conclusion on Diagonalizability

Since the number of linearly independent eigenvectors equals the size of the matrix, we conclude that the matrix $ A $ is diagonalizable.