An \((n,n)\)-matrix \(\Mtrx{A}\) is called diagonalizable if there exists an invertible matrix \(\Mtrx{C}\) such that
| \(\Mtrx{D}\) | \(=\) | \(\Mtrx{C}^{-1}\cdot \Mtrx{A} \Mtrx{C}\) |
is a diagonal matrix.
Abstract We consider the situation where a given \((n,n)\)-matrix \(\Mtrx{A}\) has a basis \(\EuScript{B}\) of eigenvectors for \(\RNrSpc{n}\). Consequently, the transformation of \(\RNrSpc{n}\) represented by \(\Mtrx{A}\) in standard coordinates is represented by a diagonal matrix using \(\mathcal{B}\)-coordinates. Accordingly, we say that \(\Mtrx{A}\) is diagonalizable. We explain how to use eigentheory to detect diagonalizable matrices and then diagonalize them.
Outline We use eigentheory techniques to detect linear transformations \(L\from \RNrSpc{n}\to \RNrSpc{n}\) which admit a basis
\[\OrdVSpcBss{B} = \left( \Vect{b}_{1,1},\dots ,\Vect{b}_{1,\AlgbrcMltplcty{\EigenVal{\lambda_1}}}, \dots , \Vect{b}_{k,1},\dots ,\Vect{b}_{k,\AlgbrcMltplcty{\EigenVal{\lambda_k}}} \right)\]of \(\RNrSpc{n}\) consisting only of eigenvectors of \(L\). Thus the associated eigenvalues are \(\lambda_1,\dots ,\lambda_k\), and the matrix which represents \(L\) using \(\mathcal{B}\)-coordinates is
\[ D \DefEq \LinMapMtrxxOnBss{L}{B}{B} = \left[ \begin{array}{ccccccc} \lambda_1 & \cdots & 0 & & 0 & \cdots & 0 \\ % \vdots & \ddots & \vdots & & \vdots && \vdots \\ % 0 & \cdots & \lambda_1 && 0 & \cdots & 0 \\ % \vdots & & \vdots & \ddots & \vdots && \vdots \\ % 0 & \cdots & 0 & & \lambda_k & \cdots & 0 \\ \vdots && \vdots & & \vdots & \ddots & \vdots \\ % 0 & \cdots & 0 & & 0 & \cdots & \lambda_k \end{array} \right] \]In other words, in \(\mathcal{B}\)-coordinates, the transformation properties of \(L\) are quite easy to understand: each coordinate axis gets rescaled by its associated eigenvalue. In this situation, we call \(L\) diagonalizable. Further, if \(\Mtrx{A}\) represents \(L\) in standard coordinates, we call \(\Mtrx{A}\) diagonalizable as well.
In fact, here is how we diagonalize \(\Mtrx{A}\). First form the coordinate conversion matrix \(\CoordTrafoMtrx{C}{S}{B}\) which converts from \(\OrdVSpcBss{B}\)-coordinates to \(\VSpcBss{S}\)-coordinates; that is:
\[ \CoordTrafoMtrx{C}{S}{B} = \left[ \begin{array}{ccccccc} \uparrow & & \uparrow && \uparrow & & \uparrow \\ % \Vect{b}_{1,1} & \cdots &\Vect{b}_{1,\AlgbrcMltplcty{\EigenVal{\lambda_1}}} & \cdots & \Vect{b}_{k,1} & \cdots & \Vect{b}_{k,\AlgbrcMltplcty{\EigenVal{\lambda_k}}} \\ % \downarrow & & \downarrow && \downarrow & & \downarrow \end{array} \right] \]Then
\[\Mtrx{A} = \CoordTrafoMtrx{C}{S}{B} \Mtrx{D} \CoordTrafoMtrx{C}{S}{B}^{-1} \quad \text{equivalently}\quad \CoordTrafoMtrx{C}{S}{B}^{-1} \Mtrx{A} \CoordTrafoMtrx{C}{S}{B} = \Mtrx{D}\]The equation on the right expresses the diagonalization of \(\Mtrx{A}\).
Detecting diagonalizable matrices How do we tell if a given matrix \(\Mtrx{A}\) is diagonalizable? – We solve this problem using eigentheory: First of all, the characteristic polynomial \(p(\lambda)\) of \(\Mtrx{A}\) must factor like this:
\[p(\lambda) = (\lambda - \EigenVal{\lambda}_1)^{\AlgbrcMltplcty{\lambda_1}} \cdots (\lambda - \EigenVal{\lambda}_k)^{\AlgbrcMltplcty{\lambda_k}},\quad \lambda_1,\dots ,\lambda_k\in\RNr.\]If \(\Mtrx{A}\) has size \((n,n)\), this implies that
\[n = \AlgbrcMltplcty{\lambda_1} + \cdots + \AlgbrcMltplcty{\lambda_k}\]Next, for each eigenvalue \(\lambda_k\) of \(\Mtrx{A}\), its algebraic multiplicity must equal its geometric multiplicity. This means that
\[n = \AlgbrcMltplcty{\lambda_1} + \cdots + \AlgbrcMltplcty{\lambda_k} = \GmtrcMltplcty{\lambda_1} + \cdots + \GmtrcMltplcty{\lambda_k}\]In other words, the dimensions of the eigenspaces \(E(\lambda_1),\dots ,E(\lambda_k)\) add up to \(n\). Therefore, the eigenspaces form a splitting of \(\RNrSpc{n}\). So, there is an ordered basis \(\mathcal{B}\) of \(\RNrSpc{n}\) consisting only of eigenvectors of \(\Mtrx{A}\). Now, we are exactly in the situation described above, and may diagonalize \(\Mtrx{A}\).
Powers of a diagonalizable matrix So far, we have focused on the geometric foundations which underlie the diagonalization of a matrix. However, there are additional reasons why one might want to diagonalize a matrix. For example, the task of computing a (high) power \(\Mtrx{A}^r\) can be quite labor intensive, especially if \(r\) is large. However, if \(\Mtrx{A} = \Mtrx{C} \Mtrx{D} \Mtrx{C}^{-1}\), with \(D\) a diagonal matrix, then
\[\Mtrx{A}^r = \Mtrx{C}\cdot \Mtrx{D}^r \cdot \Mtrx{C}^{-1}\]The point here is that \(\Mtrx{D}^r\) is again a diagonal matrix. Moreover, if \(d_i\) is the entry of \(\Mtrx{D}\) in position \((i,i)\), then \(d^{r}_{i}\) is the entry of \(\Mtrx{D}^r\) in position \((i,i)\).
Orthogonal diagonalization We emphasize one situation where diagonalizability is particularly easy to recognize, and where diagonalization is particularly easy to work with: every symmetric matrix \(\Mtrx{A}\) has an orthonormal basis of eigenvectors. Consequently, \(\Mtrx{A}\) is diagonalizable and, assuming that the basis \(\mathcal{B}\) above is an ONB,
\[\Mtrx{A} = \CoordTrafoMtrx{C}{S}{B}\cdot \Mtrx{D}\cdot \CoordTrafoMtrx{C}{S}{B}^{T} \quad \text{equivalently}\quad \CoordTrafoMtrx{C}{S}{B}^{T}\cdot \Mtrx{A}\cdot \CoordTrafoMtrx{C}{S}{B} = \Mtrx{D}\]We use the following theorem to detect diagonalizable matrices. Further, if a matrix \(\Mtrx{A}\) is diagonalizable, it provides a method to find a matrix \(\Mtrx{C}\) such that \(\Mtrx{C}^{-1} \Mtrx{A} \Mtrx{C}\) is diagonal.
An \((n,n)\)-matrix \(\Mtrx{A}\) is diagonalizable if and only if the following conditions hold:
There exist numbers \(\lambda_1,\dots ,\lambda_r\) in \(\RNr\) such that the characteristic polynomial of \(\Mtrx{A}\) is of the form
| \(p(\lambda)\) | \(=\) | \((\lambda-\lambda_1)^{a_1}\, \cdots\, (\lambda - \lambda_r)^{a_r}\) |
where \(\lambda_1,\dots ,\lambda_r\) are pairwise distinct.
Whenever these conditions are satisfied, \(\RNrSpc{n}\) has an ordered basis \(\EuScript{B} = (\Vect{b}_1 , \dots ,\Vect{b}_n)\) consisting of eigenvectors of \(\Mtrx{A}\). Let \(\Mtrx{D}\) be the \((n,n)\)-diagonal matrix whose \(i\)-th diagonal entry is the eigenvalue of \(\Vect{b}_i\), and let \(\Mtrx{C} = \CoordTrafoMtrx{C}{\EuScript{S}}{\EuScript{B}}\) be the matrix whose \(i\)-th column is \(\Vect{b}_i\). Then
| \(\Mtrx{D}\) | \(=\) | \(C^{-1}\, \Mtrx{A}\, \Mtrx{C}\) |
We have seen that diagonalizing a matrix helps us in determining the transformation properties of a given matrix. But there are other reasons why one might want to diagonalize a matrix \(\Mtrx{A}\). For example, the task of computing a (high) power \(\Mtrx{A}^r\) can be quite labor intensive, especially if \(r\) is large. The following result tells us, however, that this task can be simplified hugely if \(\Mtrx{A}\) is diagonalizable.
Let us now turn to a deeper fact: we show that every symmetric matrix \(\Mtrx{A}\) has an ONB of eigenvectors. Consequently, \(\Mtrx{A}\) is diagonalizable, and the coordinate conversion matrix may be chosen to be an orthogonal matrix.