A diagonal matrix is a square matrix that has zero entries above and below the main diagonal (the diagonal from top left to bottom right). An n x n square matrix A is diagonalizable if there exists an invertible matrix P such that P^-1AP is diagonal, which can only happen if A has n linearly independent eigenvectors, p₁, p₂, …, p_n. Then P = [p₁ p₂ … p_n] and D = P^-1AP has the eigenvalues of A on its main diagonal.

Two by two matrix example

Diagonalize matrix [latex]A=\left(\begin{array}{c}-2&3\\-4&5\end{array}\right)[/latex] if possible.

1. From Matrices: eigenvalues and eigenvectors, the eigenvalues are [latex]\lambda_1=1[/latex] and [latex]\lambda_2=2[/latex] with eigenvectors [latex]\mathbf{p}_1=\left(\begin{array}{c}1\\1\end{array}\right)[/latex] and [latex]\mathbf{p}_2=\left(\begin{array}{c}3\\4\end{array}\right)[/latex].
   Note [latex]\mathbf{p}_1[/latex] and [latex]\mathbf{p}_2[/latex] are linearly independent because the vector equation [latex]c_1\mathbf{p}_1+c_2\mathbf{p}_2=\mathbf{0}[/latex] only has the trivial solution [latex]c_1=c_2=0[/latex].
2. Then [latex]P=\left(\begin{array}{c}1&3\\1&4\end{array}\right)[/latex] and [latex]P^{-1}=\left(\begin{array}{c}4&-3\\-1&1\end{array}\right)[/latex].
3. Calculate [latex]D=P^{-1}AP[/latex][latex]=\left(\begin{array}{c}4&-3\\-1&1\end{array}\right)\left(\begin{array}{c}-2&3\\-4&5\end{array}\right)\left(\begin{array}{c}1&3\\1&4\end{array}\right)=\left(\begin{array}{c}1&0\\0&2\end{array}\right)[/latex]
   to confirm D is diagonal with the eigenvalues [latex]\lambda_1=1[/latex] and [latex]\lambda_2=2[/latex] along the main diagonal.

Three by three matrix example

Diagonalize matrix [latex]A=\left(\begin{array}{c}1&2&-1\\1&-1&2\\1&0&1\end{array}\right)[/latex] if possible.

1. From Matrices: eigenvalues and eigenvectors, the eigenvalues are [latex]\lambda_1=-1[/latex], [latex]\lambda_2=0[/latex], and [latex]\lambda_3=2[/latex] with eigenvectors [latex]\mathbf{p}_1=\left(\begin{array}{c}-4\\5\\2\end{array}\right)[/latex], [latex]\mathbf{p}_2=\left(\begin{array}{c}-1\\1\\1\end{array}\right)[/latex], and [latex]\mathbf{p}_3=\left(\begin{array}{c}1\\1\\1\end{array}\right)[/latex].
   Note [latex]\mathbf{p}_1[/latex], [latex]\mathbf{p}_2[/latex], and [latex]\mathbf{p}_3[/latex] are linearly independent because the vector equation [latex]c_1\mathbf{p}_1+c_2\mathbf{p}_2+c_3\mathbf{p}_3=\mathbf{0}[/latex] only has the trivial solution [latex]c_1=c_2=c_3=0[/latex].
2. Then [latex]P=\left(\begin{array}{c}-4&-1&1\\5&1&1\\2&1&1\end{array}\right)[/latex] and [latex]P^{-1}=\left(\begin{array}{c}0&1/3&-1/3\\-1/2&-1&3/2\\1/2&1/3&1/6\end{array}\right)[/latex].
3. Calculate [latex]D=P^{-1}AP[/latex][latex]=\left(\begin{array}{c}0&1/3&-1/3\\-1/2&-1&3/2\\1/2&1/3&1/6\end{array}\right)\left(\begin{array}{c}1&2&-1\\1&-1&2\\1&0&1\end{array}\right)\left(\begin{array}{c}-4&-1&1\\5&1&1\\2&1&1\end{array}\right)=\left(\begin{array}{c}-1&0&0\\0&0&0\\0&0&2\end{array}\right)[/latex]
   to confirm D is diagonal with the eigenvalues [latex]\lambda_1=-1[/latex], [latex]\lambda_2=0[/latex], and [latex]\lambda_3=2[/latex] along the main diagonal.