Understanding Diagonalizable Matrices

A square matrix A is called diagonalizable if it is similar to a diagonal matrix. This means there exists an invertible matrix P and a diagonal matrix D such that \(A = PDP^{-1}\).

Let's analyze each given statement about diagonalizable matrices.

Analyzing Statement 1: Every square matrix is diagonalizable

This statement claims that any square matrix, regardless of its properties, can be diagonalized. This is false.

• Consider the matrix \(A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\).
• The characteristic polynomial is \(det(A - \lambda I) = det\begin{pmatrix} -\lambda & 1 \\ 0 & -\lambda \end{pmatrix} = (-\lambda)(-\lambda) - (1)(0) = \lambda^2\).
• The eigenvalues are the roots of \(\lambda^2 = 0\), which are \(\lambda_1 = 0\) and \(\lambda_2 = 0\). The algebraic multiplicity of the eigenvalue 0 is 2.
• To find the eigenvectors for \(\lambda = 0\), we solve \((A - 0I)v = 0\), which is \(Av = 0\).
• \(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\) leads to \(y = 0\).
• The eigenvectors are of the form \(\begin{pmatrix} x \\ 0 \end{pmatrix}\) where \(x \neq 0\).
• The eigenspace for \(\lambda = 0\) is spanned by \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\). The dimension of the eigenspace (geometric multiplicity) is 1.
• Since the geometric multiplicity (1) is less than the algebraic multiplicity (2), the matrix A is not diagonalizable.

Therefore, Statement 1 is false.

Analyzing Statement 2: The sum of diagonalizable matrices is also diagonalizable

This statement claims that if you add two matrices that are diagonalizable, the resulting matrix will also be diagonalizable. This is false.

• Consider two diagonalizable matrices:
• \(A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}\). Eigenvalues are 1 and 2 (distinct), so A is diagonalizable.
• \(B = \begin{pmatrix} -1 & 0 \\ 0 & -2 \end{pmatrix}\). B is a diagonal matrix, so it is diagonalizable.
• Their sum is \(A+B = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} + \begin{pmatrix} -1 & 0 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\).
• As shown in the analysis of Statement 1, the matrix \(\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\) is not diagonalizable.

Therefore, Statement 2 is false.

Analyzing Statement 3: The product of diagonalizable matrices is also diagonalizable

This statement claims that if you multiply two matrices that are diagonalizable, the resulting matrix will also be diagonalizable. This is false.

• Consider two diagonalizable matrices:
• \(A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\). The characteristic polynomial is \(det(A - \lambda I) = det\begin{pmatrix} -\lambda & 1 \\ 1 & -\lambda \end{pmatrix} = \lambda^2 - 1\). Eigenvalues are \(\lambda = \pm 1\) (distinct), so A is diagonalizable.
• \(B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\). B is a diagonal matrix, so it is diagonalizable.
• Their product is \(AB = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\).
• The characteristic polynomial of AB is \(det(AB - \lambda I) = det\begin{pmatrix} -\lambda & 0 \\ 1 & -\lambda \end{pmatrix} = \lambda^2\). The eigenvalue is \(\lambda = 0\) with algebraic multiplicity 2.
• To find eigenvectors for \(\lambda = 0\), solve \((AB - 0I)v = 0\), which is \(ABv = 0\).
• \(\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\) leads to \(x = 0\).
• The eigenvectors are of the form \(\begin{pmatrix} 0 \\ y \end{pmatrix}\) where \(y \neq 0\).
• The eigenspace for \(\lambda = 0\) is spanned by \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\). The geometric multiplicity is 1.
• Since the geometric multiplicity (1) is less than the algebraic multiplicity (2), the matrix AB is not diagonalizable.

Therefore, Statement 3 is false.

Analyzing Statement 4: If A is a diagonalizable matrix, then A’ is diagonalizable

This statement claims that if a matrix A is diagonalizable, its transpose A’ (or A^T) is also diagonalizable. This is true.

• If A is diagonalizable, then there exists an invertible matrix P and a diagonal matrix D such that \(A = PDP^{-1}\).
• Taking the transpose of both sides, we get \(A^T = (PDP^{-1})^T\).
• Using the property \((XYZ)^T = Z^T Y^T X^T\), we have \(A^T = (P^{-1})^T D^T P^T\).
• Since D is a diagonal matrix, its transpose is itself, i.e., \(D^T = D\).
• Also, the transpose of an invertible matrix is invertible, and the inverse of the transpose is the transpose of the inverse, i.e., \((P^{-1})^T = (P^T)^{-1}\).
• Substituting these into the expression for \(A^T\), we get \(A^T = (P^T)^{-1} D P^T\).
• Let \(Q = P^T\). Since P is invertible, Q is also invertible.
• So, we have \(A^T = Q^{-1} D Q\).
• This equation shows that \(A^T\) is similar to the diagonal matrix D, with the invertible matrix Q.
• Therefore, \(A^T\) is diagonalizable.

Therefore, Statement 4 is true.

Based on the analysis, only Statement 4 is true.

Revision Table: Summary of Diagonalizability Properties

Additional Information on Diagonalizable Matrices

A matrix being diagonalizable is a crucial concept in linear algebra, linked to its eigenvalues and eigenvectors.

• Definition: A square matrix A is diagonalizable if and only if there exists an invertible matrix P such that \(P^{-1}AP\) is a diagonal matrix.
• Eigenvalue Condition: A matrix is diagonalizable if and only if the sum of the dimensions of its eigenspaces equals the size of the matrix. For each eigenvalue, the geometric multiplicity must equal the algebraic multiplicity.
• Distinct Eigenvalues: If a matrix has n distinct eigenvalues (where n is the size of the matrix), then it is guaranteed to be diagonalizable.
• Geometric Interpretation: Diagonalizability means that the vector space has a basis consisting entirely of eigenvectors of the matrix.
• Applications: Diagonalization simplifies matrix powers \(A^k = PD^kP^{-1}\), solving systems of linear differential equations, and understanding the behavior of linear transformations.
• Commuting Matrices: If two diagonalizable matrices commute (AB = BA), then they are simultaneously diagonalizable, meaning there exists a single invertible matrix P such that both \(P^{-1}AP\) and \(P^{-1}BP\) are diagonal. This is related to the sum/product properties, as commutativity is often required for those properties to hold generally.