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Wize University Linear Algebra Textbook > Eigenvalues and Eigenvectors (Spectral Theory)

Diagonalization

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Diagonalization

A matrix AA is diagonalizable if AA is similar to a diagonal matrix DD, that is, D=P−1APD=P^{-1}AP.

Diagonalization Theorem

An×nA_{n\times n} is diagonalizable if and only if AA has nn linearly independent eigenvectors: x⃗1,x⃗2,…,x⃗n\vec x_1, \vec x_2, \dots, \vec x_n.
How to Diagonalize A\colorThree A
  • Use the eigenvectors of AA as the columns of PP:
P=[∣∣∣x⃗1x⃗2…x⃗n∣∣∣]P = \left[\begin{array}{ccc} | & | & & |\\ \vec x_1 & \vec x_2 & \dots & \vec x_n\\ | & | & & |\\ \end{array} \right]
  • The diagonal entries of DD are the eigenvalues of AA (in the same order as the eigenvectors in PP):
D=[λ10000λ20000⋱0000λn]D = \left[ \begin{array}{rrr} \lambda_1 & 0 & 0 & 0\\ 0 & \lambda_2 & 0& 0\\ 0 & 0 & \ddots & 0\\ 0 & 0 & 0 & \lambda_n\\ \end{array} \right]

Wize Tip
AA is diagonalizable if the algebraic multiplicity matches the geometric multiplicity for all eigenvalues.
In particular, if AA has nn distinct eigenvalues, then AA is diagonalizable.

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Powers of Matrices

Diagonal Matrices
If DD is a diagonal matrix, then DkD^k can be found by raising every diagonal entry to the power of kk.
D=[d10⋯00d2⋯0⋮⋮⋱⋮00⋯dn]  ⟹  Dk=[d1k0⋯00d2k⋯0⋮⋮⋱⋮00⋯dnk]D=\left[\begin{array}{ccccc} d_1&0&\cdots&0\\[0.5em] 0&d_2&\cdots&0\\[0.5em] \vdots&\vdots&\ddots&\vdots\\[0.5em] 0&0&\cdots&d_n \end{array}\right] \quad \implies \quad D^{\colorTwo{k}}=\left[\begin{array}{ccccc} \colorTwo{d_1^k}&0&\cdots&0\\[0.5em] 0&\colorTwo{d_2^k}&\cdots&0\\[0.5em] \vdots&\vdots&\ddots&\vdots\\[0.5em] 0&0&\cdots&\colorTwo{d_n^k} \end{array}\right]
Other Matrices
We can use diagonalizability to compute large powers of any diagonalizable matrix AA.
Since AA is similar to a diagonal matrix DD, we can write D=P−1APD = P^{-1}AP. Solve for AA:
P−1AP=DPP−1APP−1=PDP−1A=PDP−1\begin{aligned} P^{-1}AP &= D\\ \colorOne{P}P^{-1}AP\colorTwo{P^{-1}} &= \colorOne{P}D\colorTwo{P^{-1}}\\ A &= \colorOne{P}D\colorTwo{P^{-1}} \end{aligned}
We can use this expression for AA to compute AkA^k:
Ak=(PDP−1)k=(PDP−1)(PDP−1)⋯(PDP−1)=PD  P−1P  D  P−1P  ⋯P−1P  DP−1=PDD⋯DP−1=PDkP−1\begin{array}{rcl} A^k&=&(PDP^{-1})^k\\[1em] &=& (PDP^{-1})(PDP^{-1})\cdots(PDP^{-1})\\[1em] &=& PD\;\colorTwo{P^{-1}P}\;D\;\colorTwo{P^{-1}P}\;\cdots \colorTwo{P^{-1}P}\;DP^{-1}\\[1em] &=& PDD\cdots DP^{-1}\\[1em] &=& PD^kP^{-1} \end{array}

Ak=PDkP−1\boxed{\quad A^k = PD^kP^{-1} \quad}

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Example: Diagonalization and Matrix Powers

Let A=[−34−23]A=\left[\begin{array}{rr} -3&4\\-2&3\end{array}\right].

Part A)

Find an invertible matrix PP and a diagonal matrix DD such that AA is similar to DD.
Eigenvalues
det(A−λI)=0  ⟹  ∣−3−λ4−23−λ∣=0(−3−λ)(3−λ)−4(−2)=0λ2−1=0\begin{array}{rcr} \text{det}(A-\lambda I)=0 &\implies& \left|\begin{array}{cc} -3-\lambda&4\\-2&3-\lambda\end{array}\right|=0\\[1.5em] && (-3-\lambda)(3-\lambda)-4(-2)=0\\[1em] &&\lambda^2-1=0 \end{array}
The eigenvalues are λ1=−1\colorOne{\lambda_1=-1} and λ2=1\colorTwo{\lambda_2=1}.
Note that since AA is 2×22\times 2 and has 22 distinct eigenvalues, AA is diagonalizable.
Eigenvectors
When λ1=−1\colorOne{\lambda_1=-1}, we solve the linear system: (A−(−1)I)x⃗1=0⃗(A-(-1)I) \colorOne{\vec x_1} = \vec 0 whose augmented matrix is:
[−3−(−1)40−23−(−1)0]=[−240−240]  ⟹  [1−20000]\left[\begin{array}{cc|r} -3-(-1)&4&0\\-2&3-(-1)&0\end{array}\right] = \left[\begin{array}{cc|r} -2&4&0\\-2&4&0\end{array}\right] \implies \left[\begin{array}{rr|r} 1&-2&0\\0&0&0\end{array}\right]
The solution to this system is any vector of the form:
[2tt]=t[21]\left[\begin{array}{r} 2t\\t\end{array}\right] = t\left[\begin{array}{r} 2\\1\end{array}\right]
So the eigenvector is: x⃗1=[21]\colorOne{\vec x_1 = \left[\begin{array}{r} 2\\1\end{array}\right]}
When λ2=1\colorTwo{\lambda_2=1}, we solve the linear system: (A−1I)x⃗2=0⃗(A-1I)\colorTwo{\vec x_2} = \vec 0 whose augmented matrix is:
[−3−140−23−10]=[−440−220]  ⟹  [1−10000]\left[\begin{array}{cc|r} -3-1&4&0\\-2&3-1&0\end{array}\right] = \left[\begin{array}{cc|r} -4&4&0\\-2&2&0\end{array}\right] \implies \left[\begin{array}{rr|r} 1&-1&0\\0&0&0\end{array}\right]
The solution to this system is any vector of the form: [tt]=t[11]\left[\begin{array}{r} t\\t\end{array}\right] = t\left[\begin{array}{r} 1\\1\end{array}\right]
So the eigenvector is: x⃗2=[11]\colorTwo{\vec x_2 = \left[\begin{array}{r} 1\\1\end{array}\right]}
Then:
P=[2111]P=\left[\begin{array}{rr} \colorOne{2}&\colorTwo{1}\\ \colorOne{1}&\colorTwo{1} \end{array}\right] which means P−1=[1−1−12]P^{-1}=\left[\begin{array}{rr} 1&-1\\-1&2\end{array}\right], and AA is similar to D=[−1001]D=\left[\begin{array}{rr} \colorOne{-1}&0\\ 0&\colorTwo{1} \end{array}\right].
Note the order of the eigenvectors in PP and their associated eigenvalues in DD.

Part B)

Compute A100A^{100}.
Using the diagonalization of AA:
A100=(PDP−1)100=PD100P−1=[2111][−1001]100[1−1−12]=[2111][(−1)100001100][1−1−12]=[2111][1001]⏟[1−1−12]=[2111][1−1−12]=[1001]\begin{array}{rcl} A^{100}&=&(PDP^{-1})^{100}\\[1em] &=&PD^{100}P^{-1}\\[1em] &=& \left[\begin{array}{rr} 2&1\\1&1\end{array}\right] \left[\begin{array}{rr} -1&0\\0&1\end{array}\right]^{100} \left[\begin{array}{rr} 1&-1\\-1&2\end{array}\right]\\[2em] &=& \left[\begin{array}{rr} 2&1\\1&1\end{array}\right] \left[\begin{array}{cc} (-1)^{100}&0\\0&1^{100}\end{array}\right] \left[\begin{array}{rr} 1&-1\\-1&2\end{array}\right]\\[2em] &=& \underbrace{ \left[\begin{array}{rr} 2&1\\1&1\end{array}\right] \left[\begin{array}{cc} 1&0\\0&1\end{array}\right]} \left[\begin{array}{rr} 1&-1\\-1&2\end{array}\right]\\[2em] &=& \left[\begin{array}{rr} 2&1\\1&1\end{array}\right] \left[\begin{array}{rr} 1&-1\\-1&2\end{array}\right]\\[2em] &=& \boxed{ \left[\begin{array}{rr} 1&0\\0&1\end{array}\right]} \end{array}

Practice: Determining Diagonalizability

Let A=[−79−35]A=\left[\begin{array}{rr}-7&9\\-3&5\end{array}\right]. Is AA diagonalizable?

Practice: Diagonalization

Let A=[260−1]A=\left[\begin{array}{rr} 2&6\\ 0&-1\end{array}\right].
First, convince yourself (without calculation) that AA is diagonalizable.
Then, compute A50A^{50}.
Extra Practice
Diagonalization
Given that the matrix A=[−32611−1942−14−6]A=\left[\begin{array}{rrr}-3&26&11\\-1&9&4\\2&-14&-6\end{array}\right] has characteristic polynomial (λ+1)2(λ−2)(\lambda+1)^2(\lambda-2) and the matrix B=[−12300−513010−30−2]B=\left[\begin{array}{rrr}-12&30&0\\-5&13&0\\10&-30&-2\end{array}\right] has characteristic polynomial (λ+2)2(λ−3)(\lambda+2)^2(\lambda-3) , determine which of A,BA,B are diagonalizable, and in the case that they are diagonalizable, find the diagonal matrix DD and the diagonalizing matrix PP.
133 - FML 3 - 18.1W e.g. 76
Find the matrices   ⁣P‾\bcb{\boldsymbol{ \ul{P}}} and   ⁣D‾\bcb{\boldsymbol{ \ul{D}}} that diagonalize the matrix   ⁣A‾=[10413−2101]\bcb{\boldsymbol{ \ul{A } = \begin{bmatrix} 1 & 0 & 4 \\ 1 & 3 & -2 \\ 1 & 0 & 1 \end{bmatrix} }}, if they exist
133 - FML 3 - 18.1W e.g. 73
Find the matrices   ⁣P‾\bcb{\boldsymbol{ \ul{P}}} and   ⁣D‾\bcb{\boldsymbol{ \ul{D}}} that diagonalize the matrix   ⁣A‾=[4−162162−18]\bcb{\boldsymbol{ \ul{A } = \begin{bmatrix} 4 & -1 & 6 \\ 2 & 1 & 6 \\ 2 & -1 & 8 \end{bmatrix} }}, if they exist
.
133 - FML 3 - 18.1W e.g. 72
Find the matrices   ⁣P‾\bcb{\boldsymbol{ \ul{P}}} and   ⁣D‾\bcb{\boldsymbol{ \ul{D}}} that diagonalize the matrix   ⁣A‾=[3−20−230005]\bcb{\boldsymbol{ \ul{A } = \begin{bmatrix} 3 & -2 & 0 \\ -2 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} }}, if they exist
.
133 - FML 3 - 18.1W e.g. 68
Find all value of k\bcb{\boldsymbol{ k}} such that the matrix   ⁣M‾=[1k02]\bcb{\boldsymbol{ \ul{M} = \begin{bmatrix} 1 & k \\ 0 & 2 \end{bmatrix} }} is fully diagonalizable.
Let
A=[−2273]A=\begin{bmatrix}-2&2\\7&3\end{bmatrix}
Find an invertible matrix PPand a diagonal matrix DD such that P−1AP=DP^{-1}AP = D, and thus find A10A^{10}.
Practice Question: Diagonalizing a Matrix
Is the following matrix diagonalizable? If so, find a diagonalization for it.
A=(−1522)A = \left ( \begin{array}{cc} -1 & 5 \\ 2 & 2 \end{array} \right )
Diagonalization
Practice: Diagonalization
Let A=[260−1]A=\left[\begin{array}{rr} 2&6\\ 0&-1\end{array}\right].
First, convince yourself (without calculation) that AA is diagonalizable.
Properties
Practice: Determining Diagonalizability
Let A=[−79−35]A=\left[\begin{array}{rr}-7&9\\-3&5\end{array}\right]. Is AA diagonalizable?
133 Mid 2 18.4F e.g. 15
If A‾=[−8k0−8]\A=\sm {-8}{k}{0}{-8}, then determine the values of k for which A‾\A is diagonalizable.
Enter the values of k in increasing order, separated by a comma.
For example of you find k = -2, k = 1 and k = 0, enter your answer as "-2, 0, 1". If there are none, enter the ̸ ⁣∃\not\!\exist symbol (below the equation palette).
19.4F_Final_Builder_Ch_12.7_Dynamical_Systems_$\tkcth{eg7}$_$\key{Final}$_Builder_$\tkcth{12.7.}\tkcf{7}$_
A basis for R2\mathbb{R}^2 is given by the set E={ξ⃗1, ξ⃗2}E = \left\{\vXi_1,\, \vXi_2 \right\}, where ξ⃗1=[31]\vXi_1 = \colvec{3}{1} and ξ⃗2=[12]\vXi_2 = \colvec{1}{2} are the eigenvectors of the matrix
A‾=[53−212] \A = \sm{5}{3}{-2}{12}
Find [x⃗]E[ \vx ]_{E}, the coordinates (in the EE basis) of the vector whose standard coordinates are x⃗=[1−1]\vx = \colvec{1}{-1} .
133 - FML 1 - 18.1W e.g. 10.1
A square n × n matrix AA is said to be diagonalizable if it is similar to a diagonal matrix, i.e. A‾=P ⁣‾ D ⁣‾ P ⁣‾ −1\A = \P \bct{\D} \P^{\bct{-1}} .
Rearranging, we have the definition that A‾\A is diagonlizable if D ⁣‾ =P ⁣‾ −1A‾P ⁣‾ \D=\P^{\bct{-1}} \A \P is diagonal.
The notation itself should make clear that this is only possible if there is an n×nn\times n matrix P ⁣‾ \P that is invertible (i.e. ∃  P ⁣‾ −1  :  P ⁣‾ P ⁣‾ −1  =  I ⁣‾ \exist \; \minv{\P} \; : \; \P\minv{\P} \; = \;\I). The question then arises: what is P ⁣‾ \P and how can we find it?
133 - FML 1 - 18.1W e.g. 9
If B ⁣‾ = diag(λ1,λ2,λ3)\B=\text{ diag}(\lambda_1,\lambda_2,\lambda_3) find an expression for the matrix-matrix multiplication:
[x⃗1x⃗2x⃗3]B ⁣‾  \big[{\vx}_1\quad{\vx}_2\quad {\vx}_3\big]\B,
in terms of the diagonal entries λi\lambda_i of B ⁣‾ \B, and the 3×13 \times 1column vectors x⃗i\vx_i.
Practice Question: Diagonalization
Consider the matrix A=[2014]A=\begin{bmatrix}2 & 0\\1&4\end{bmatrix}. Which of the following matrices PP diagonalizes AA?