Eigenvalues describe individual special directions. Diagonalization asks whether there are enough independent eigenvectors to form a basis. In eigenvector coordinates, the matrix acts by scaling each coordinate separately.
An \(n\times n\) matrix \(A\) is called diagonalizable if there exists an invertible \(n \times n\) matrix \(P\) so that \(P^{-1}AP\) is a diagonal matrix (recall Definition 2.5.36). The matrix \(P\) is called a diagonalizing matrix for \(A\text{.}\)
The previous activity shows why diagonalizing a matrix may be useful. But given a matrix \(A\text{,}\) how do we potentially find \(P\) such that \(P^{-1}AP\) is a diagonal matrix?
Activity5.2.3.Interpreting the columns in a diagonalization (U5-LO1, U5-LO2).
Suppose \(P^{-1}AP = D\) for some diagonal matrix \(D = \diag(\lambda_1,\dots,\lambda_n)\text{.}\) Let \(\mathbf{v}_1, \dots, \mathbf{v}_n\) be the columns of \(P\text{.}\) Note that the equation \(P^{-1} A P = D\) holds if and only if \(AP = PD\text{.}\)
Calculate \(AP\) and \(PD\) in terms of the vectors \(\mathbf{v}_1,\dots,\mathbf{v}_n\text{,}\) the scalars \(\lambda_1,\dots,\lambda_n\) and the matrix \(A\text{.}\)
For the equation \(AP = PD\) to hold, we need \(A \mathbf{v}_i = \lambda_i \mathbf{v}_i\) for each \(i\text{.}\) Thus we need each column of \(P\) to be an eigenvector of \(A\text{,}\) whose eigenvalue is the corresponding entry of the diagonal matrix \(D\text{.}\)
To see this differently, suppose \(P^{-1}AP\) is a diagonal matrix with diagonal entries \(\lambda_1,\dots,\lambda_n\text{.}\) Then for each standard basis vector (recall Definition 1.3.14) \(P^{-1}AP \mathbf{e}_i = \lambda_i \mathbf{e}_i\text{.}\) Multiplying this equation by \(P\) on the left-hand side and setting \(\mathbf{v}_i = P \mathbf{e}_i\text{,}\) we obtain
An \(n \times n\) matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) eigenvectors \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\) such that the matrix \(P\) with columns \(\mathbf{v}_1,\dots,\mathbf{v}_n\) is invertible. In this case, \(P^{-1}AP=D\text{,}\) where \(D\) is a diagonal matrix such that the elements of \(D\) on the main diagonal are the eigenvalues of \(A\text{.}\)
Recalling Theorem 2.5.19, the matrix \(P\) in Theorem 5.2.4 is invertible if and only if the vectors \(\mathbf{v}_1,\dots,\mathbf{v}_n\) form a basis for \(\R^n\text{.}\)
If \(\R^n\) has a basis of eigenvectors \(\mathbf{v}_1,\ldots,\mathbf{v}_n\) with corresponding eigenvalues \(\lambda_1,\ldots,\lambda_n\text{,}\) then every vector \(\mathbf{v} \in \R^n\) can be written as a linear combination of these eigenvectors, i.e., there exists constants \(a_1,\dots,a_n\) such that
Therefore, \(\mathbf{u}\) is an eigenvector with eigenvalue \(\lambda_1 = 1\) and \(\mathbf{v}\) is an eigenvector with eigenvalue \(\lambda_2 = 2\text{.}\)
Figure 5.2.7 illustrates the effect of the matrix \(M\) from Activity 5.2.6 on a grid in \(\R^2\text{.}\) In particular, it highlights an important geometric interpretation of eigenvectors and eigenvalues; when we apply the linear map defined by the matrix \(M\) to any vector in \(\R^2\text{,}\) the matrix map stretches eigenvector directions by their eigenvalues.
The left panel shows a square coordinate grid with a red line in the direction \(\mathbf{u}=(1,1)\) and a blue line in the direction \(\mathbf{v}=(1,-1/2)\text{.}\) The right panel shows the image of the grid under \(M\text{.}\) The red direction is unchanged, while the blue direction is stretched to twice its original length.
Figure5.2.7.Visualization of the action of the matrix \(M\) from Activity 5.2.6 on a grid. The red line along the eigenvector \(\mathbf{u}\) is unchanged, because \(\mathbf{u}\) has eigenvalue \(1\text{.}\) The blue line along the eigenvector \(\mathbf{v}\) is stretched by a factor of \(2\text{,}\) because \(\mathbf{v}\) has eigenvalue \(2\text{.}\) Adapted from: Stanford’s MATH 51 textbook.
where \(\theta\) is not a multiple of \(180°\text{.}\) For any nonzero vector \(\mathbf{v} \in \R^2\text{,}\) the vector \(A_\theta\mathbf{v}\) is obtained by rotating \(\mathbf{v}\) counterclockwise by an angle \(\theta\text{,}\) so \(A_\theta\mathbf{v}\) is never on the line spanned by \(\mathbf{v}\text{.}\) Thus \(A_\theta\mathbf{v}=\lambda\mathbf{v}\) cannot hold for any real scalar \(\lambda\) and any nonzero \(\mathbf{v}\text{.}\)
Thus the matrix \(A_\theta\) has no real eigenvalues or real eigenvectors. It is therefore not diagonalizable over \(\R\text{.}\) This makes geometric sense: we cannot find special directions in \(\R^2\) so that rotating vectors in the plane is obtained by stretching in those directions.
Remark 5.2.8 illustrates why not every matrix is diagonalizable. A matrix is diagonalizable precisely when it has a basis of eigenvectors, but this is not always possible.
Recalling Fact 5.1.11, every eigenvector of \(A\) is a scalar multiple of this vector. So \(A\) cannot have a basis of eigenvectors, and so \(A\) is not diagonalizable.
