Exercise · ZW-DIAGONALIZATION-005 · ★★☆ Intermediate
Let \(\begin{pmatrix}2&1\\0&2\end{pmatrix}\). Is it diagonalizable?
Choose one option.
Show answer
- C. No: it has only one independent eigenvector
Explanation
The only eigenvalue is 2 and \(\dim E_2=1<2\); there is no basis of eigenvectors. Being triangular is not enough.
Review the concept: Diagonalization
A square matrix \(A\) is diagonalizable if it is similar to a diagonal matrix \(D\), that is, if it can be written as an invertible matrix \(P\) times \(D\) times the inverse of \(P\).
\[A=PDP^{-1}\]
More practice problems: diagonalization
- ZW-DIAGONALIZATION-001 Let [4 1; 2 3] with eigenvalues 5 and 2. With D=diag(5,2), which matrix P satisfies P^-1AP=D?
- ZW-DIAGONALIZATION-002 A 2× 2 matrix has eigenvalues 1 and 4. Is it diagonalizable?
- ZW-DIAGONALIZATION-003 A 4× 4 matrix is diagonalizable. How many linearly independent eigenvectors does it have?
- ZW-DIAGONALIZATION-004 A diagonalizable matrix A=PDP^-1 has D=diag(2,-1). What is the diagonal matrix of A^3 in the same b…
- ZW-DIAGONALIZATION-006 A diagonalizable matrix has trace 7 and determinant 10. Which diagonal matrix is similar to it?
- ZW-DIAGONALIZATION-007 With A=[4 1; 2 3], A=PDP^-1 and D=diag(5,2), what is the diagonal matrix of A^5?
- ZW-DIAGONALIZATION-008 Let [1 1 0; 0 1 0; 0 0 2], with eigenvalues 1 (double) and 2. Is it diagonalizable?
- ZW-DIAGONALIZATION-009 A 3× 3 matrix has eigenvalues 2 (double) and 5 (simple). Which condition is equivalent to being dia…