Xm3, Xm4 - Diagonalisation and matrix powers

Xm3, Xm4 - Diagonalisation and matrix powers

Use a matrix of eigenvectors to put a matrix with distinct real eigenvalues into diagonal form. Then calculate powers of matrices of order 22 and 33, keeping each eigenvector matched to its eigenvalue and stating when negative powers exist.

Put eigenvectors in columns

How could you find the fifth power of (3113)\begin{pmatrix}3&1\\1&3\end{pmatrix} without multiplying five copies?

Along (11)\begin{pmatrix}1\\1\end{pmatrix}, each application multiplies by 44; along (1−1)\begin{pmatrix}1\\-1\end{pmatrix}, it multiplies by 22. Five applications should therefore multiply those components by 454^5 and 252^5. We need a way to express an arbitrary vector using those directions.

You already know

Find eigenvalues from det⁡(A−λI)=0\det(\mathbf{A}-\lambda\mathbf{I})=0, then solve (A−λI)v=0(\mathbf{A}-\lambda\mathbf{I})\mathbf{v}=\mathbf{0}. Matrix multiplication is associative, and P−1P=I\mathbf{P}^{-1}\mathbf{P}=\mathbf{I} when P\mathbf{P} is non-singular.

For distinct real eigenvalues, the associated eigenvectors are independent. In order 22 or 33, collecting one eigenvector for each eigenvalue therefore gives an invertible matrix. We use this distinct-real case throughout the lesson.

Matrix of eigenvectors and diagonal form

Put the eigenvectors of A\mathbf{A} in the columns of P\mathbf{P}, and put their associated eigenvalues in the same order down the diagonal of D\mathbf{D}. For example,

P=(111−1),D=(4002)\mathbf{P}=\begin{pmatrix}1&1\\1&-1\end{pmatrix}, \qquad \mathbf{D}=\begin{pmatrix}4&0\\0&2\end{pmatrix}

matches the first column to 44 and the second to 22. The diagonal entries come from the eigenvalues of A\mathbf{A}, not those of P\mathbf{P}.

Multiplication by A\mathbf{A} acts separately on every column of P\mathbf{P}. Thus AP\mathbf{A}\mathbf{P} has columns λ1v1,λ2v2\lambda_1\mathbf{v}_1,\lambda_2\mathbf{v}_2, and, in three dimensions, λ3v3\lambda_3\mathbf{v}_3. This is exactly PD\mathbf{P}\mathbf{D}.

Diagonalisation

For a matrix of order 22 or 33 with distinct real eigenvalues, construct P\mathbf{P} and D\mathbf{D} in matching order. Then

AP=PD,P−1AP=D,A=PDP−1\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}, \qquad \mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{D}, \qquad \mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}

The middle equation reduces A\mathbf{A} to diagonal form.

Not in the formula booklet: learn it.

A calculation of AP\mathbf{A}\mathbf{P} checks the column matching without multiplying three matrices.

Worked example 1: diagonalising a symmetric matrix

Question. Reduce A=(3113)\mathbf{A}=\begin{pmatrix}3&1\\1&3\end{pmatrix} to diagonal form, using eigenvectors (11)\begin{pmatrix}1\\1\end{pmatrix} and (1−1)\begin{pmatrix}1\\-1\end{pmatrix}, in that order.

Plan. Identify the eigenvalue for each column, find P−1\mathbf{P}^{-1}, and verify the diagonal relation.

P=(111−1),D=(4002)match the columnsdet⁡P=−2,P−1=12(111−1)inverse formulaAP=(424−2)=PDcolumn checkP−1AP=12(111−1)(424−2)=(4002)diagonal form\begin{aligned} \mathbf{P}&=\begin{pmatrix}1&1\\1&-1\end{pmatrix}, &\mathbf{D}&=\begin{pmatrix}4&0\\0&2\end{pmatrix} &&\text{match the columns}\\ \det\mathbf{P}&=-2, &\mathbf{P}^{-1}&=\frac12\begin{pmatrix}1&1\\1&-1\end{pmatrix} &&\text{inverse formula}\\ \mathbf{A}\mathbf{P}&=\begin{pmatrix}4&2\\4&-2\end{pmatrix} & &=\mathbf{P}\mathbf{D} &&\text{column check}\\ \mathbf{P}^{-1}\mathbf{A}\mathbf{P} &=\frac12\begin{pmatrix}1&1\\1&-1\end{pmatrix} \begin{pmatrix}4&2\\4&-2\end{pmatrix} & &=\begin{pmatrix}4&0\\0&2\end{pmatrix} &&\text{diagonal form} \end{aligned}

Why this step? The eigenvectors change the coordinate description, so the matrix becomes two separate scalar multipliers.

Answer. P−1AP=diag⁡(4,2)\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\operatorname{diag}(4,2), with P\mathbf{P} as above.

Check. Multiplying PP−1\mathbf{P}\mathbf{P}^{-1} gives I\mathbf{I}.

Change only the order of the eigenvector columns. The diagonal entries must follow them.

Worked example 2: the same matrix, a different column order

Question. Diagonalise the same A\mathbf{A} with Q=(11−11)\mathbf{Q}=\begin{pmatrix}1&1\\-1&1\end{pmatrix}.

Plan. Read the first column's eigenvalue first, then calculate the transformed matrix.

D′=(2004)reversed eigenvalue orderQ−1=12(1−111)determinant 2AQ=(24−24)apply to each columnQ−1AQ=12(1−111)(24−24)=(2004)matching diagonal\begin{aligned} \mathbf{D}'&=\begin{pmatrix}2&0\\0&4\end{pmatrix} &&\text{reversed eigenvalue order}\\ \mathbf{Q}^{-1}&=\frac12\begin{pmatrix}1&-1\\1&1\end{pmatrix} &&\text{determinant }2\\ \mathbf{A}\mathbf{Q}&=\begin{pmatrix}2&4\\-2&4\end{pmatrix} &&\text{apply to each column}\\ \mathbf{Q}^{-1}\mathbf{A}\mathbf{Q} &=\frac12\begin{pmatrix}1&-1\\1&1\end{pmatrix} \begin{pmatrix}2&4\\-2&4\end{pmatrix} =\begin{pmatrix}2&0\\0&4\end{pmatrix} &&\text{matching diagonal} \end{aligned}

Why this step? Eigenvectors have no compulsory order, but a choice of column order fixes the diagonal order.

Answer. A=Qdiag⁡(2,4)Q−1\mathbf{A}=\mathbf{Q}\operatorname{diag}(2,4)\mathbf{Q}^{-1}.

Check. The first column is multiplied by 22, the second by 44, exactly as the displayed AQ\mathbf{A}\mathbf{Q} shows.

You may also rescale a column by any non-zero constant; recalculate the inverse if you do. Repeating a parallel column instead of using independent eigenvectors would make the column matrix singular.

Watch out

Wrong: put eigenvectors in rows, or reorder the columns while leaving D\mathbf{D} unchanged. Right: columns of eigenvectors, matching diagonal entries. Check AP=PD\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}.

Keep the same eigenvector directions and change the matrix's two scalar multipliers.

Raise the diagonal entries to the power

Multiplying two copies of PDP−1\mathbf{P}\mathbf{D}\mathbf{P}^{-1} brings the middle inverse and matrix together:

A2=PDP−1P⏟IDP−1=PD2P−1\mathbf{A}^2 =\mathbf{P}\mathbf{D} \underbrace{\mathbf{P}^{-1}\mathbf{P}}_{\mathbf{I}} \mathbf{D}\mathbf{P}^{-1} =\mathbf{P}\mathbf{D}^2\mathbf{P}^{-1}

The same cancellation continues for every positive integer power.

Powers from diagonal form

If A=PDP−1\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}, then

An=PDnP−1,Dn=diag⁡(λ1n,λ2n)\mathbf{A}^n=\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}, \qquad \mathbf{D}^n=\operatorname{diag}(\lambda_1^n,\lambda_2^n)

with a third diagonal entry in order 33. This holds for positive integers nn; A0=I\mathbf{A}^0=\mathbf{I}. Negative integer powers also exist when every eigenvalue is non-zero.

Not in the formula booklet: learn it.

This is the useful surprise: a large matrix power needs only powers of two or three numbers. The conversion matrices are used once at either end.

[DIAGRAM: asset_name: Coordinates along eigenvectors; asset_slug: eigenbasis_power_route; caption: Change coordinates once, scale independently, then change back; description: Four widely spaced cards show a physical vector3,1, its coordinates2,1 along the columns of P, the scaled coordinates8,2 after D=diag4,2, and the physical image10,6. Connecting arrows are labelled P inverse, D and P. Below, an n-step route replaces only D by D to the power n, keeping the two conversion matrices unchanged.]
Change coordinates once, scale independently, then change back

For the opening matrix, multiplication of the three factors gives the reusable expression

An=12(4n+2n4n−2n4n−2n4n+2n)\mathbf{A}^n =\frac12\begin{pmatrix} 4^n+2^n&4^n-2^n\\ 4^n-2^n&4^n+2^n \end{pmatrix}

This is a consequence for this particular matrix, rather than another general formula to memorise.

Continue this lesson.

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