Mathematics Prelims

June 25, 2009

Eigenvalues and Eigenvectors

Filed under: Algebra,Linear Algebra — cjohnson @ 2:39 pm

Let’s suppose that A is an n \times n matrix which is similar to a diagonal matrix, \text{diag}(\lambda_1, \lambda_2, ..., \lambda_n).  This means there is an invertible (change-of-basis) matrix P such that

\displaystyle A = P \text{diag}(\lambda_1, ..., \lambda_n) P^{-1}

Now since P is a change of basis matrix, each of its columns gives the coordinates to a basis vector of some basis.  Let’s call that basis \beta and let \beta_1 through \beta_n be the elements of that basis.  Now, if we take the above equation and multiply by P on the right, notice that

AP = P \text{diag}(\lambda_1, ..., \lambda_n)

\implies (AP)_{*i} = (P \text{diag}(\lambda_1, ..., \lambda_n))_{*i} = \lambda_i P_{*i}

That is, the i-th column of AP is equal to the i-th column of P \text{diag}(\lambda_1, ..., \lambda_n), which is just \lambda_i times the i-th column of P.  Since each column of AP is just a linear combination of the columns of A, though, we have

AP_{*i} = \lambda_i P_{*i}

This means that when we plug in the i-th column of P to the linear transformation represented by A, we get back a multiple of that column.  Calling the linear transformation \tau, we have that

\tau(\beta_i) = \lambda_i \beta_i.

Vectors such as \beta_i whose image under \tau is just a multiple of the vector are called eigenvectors of \tau.  That multiple, the \lambda_i above, is called an eigenvalue of \tau.  These eigenvectors and eigenvalues are associated with a particular linear transformation, so when we talk about the eigenvectors and eigenvalues of a matrix, we really mean the eigenvectors and eigenvalues of the transformation represented by that matrix.  Notice that this means that eigenvalues are independent of the chosen basis; since similar matrices represent the same transformation just with respect to different bases, similar matrices have the same eigenvalues.

We assumed that A was similar to a diagonal matrix above, but this isn’t always true.  If A is similar to a diagonal matrix, say A = P^{-1}DP, then as we’ve just shown, the columns of P are eigenvectors of A.  Since these form the columns of a non-singular matrix, the eigenvectors of A form a basis for the vector space.  Also, if the eigenvectors of A form a basis, let’s take those basis vectors as columns of P.

\displaystyle P^{-1} A P

\displaystyle = \left[ \begin{array}{c|c|c} \beta_1 & ... & \beta_n \end{array} \right]^{-1} A \left[ \begin{array}{c|c|c} \beta_1 & ... & \beta_n \end{array} \right]

\displaystyle = \left[ \begin{array}{c|c|c} \beta_1 & ... & \beta_n \end{array} \right]^{-1} \left[ \begin{array}{c|c|c} A \beta_1 & ... & A \beta_n \end{array} \right]

\displaystyle = \left[ \begin{array}{c|c|c} \beta_1 & \cdots & \beta_n \end{array} \right]^{-1} \left[ \begin{array}{c|c|c} \lambda_1 \beta_1 & \cdots & \lambda_n \beta_n \end{array} \right]

\displaystyle = \left[ \begin{array}{c|c|c} \beta_1 & \cdots & \beta_n \end{array} \right]^{-1} \left( \left[ \begin{array}{c|c|c} \beta_1 & \cdots & \beta_n \end{array} \right] \left[ \begin{array}{cccc} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{array} \right] \right)

\displaystyle = \text{diag}(\lambda_1, ..., \lambda_n)

So a matrix is diagonalizable (similar to a diagonal matrix) if and only if its eigenvectors form a basis for the vector space.

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4 Comments »

  1. So can I say from your post that an n\times n matrix A is diagonalizable if and only if there are linearly independent eigent vectors $latex\{\beta_1,\cdots,\beta_n\}$ of A.

    Comment by watchmath — June 25, 2009 @ 9:38 pm | Reply

  2. Right. If you have n linearly independent eigenvectors, just take those to be the columns of your P matrix. When you multiply out P^{-1} A P, you’ll get a diagonal matrix with the eigenvalues of A on the diagonal.

    Comment by cjohnson — June 25, 2009 @ 10:16 pm | Reply

  3. […] Filed under: Algebra, Linear Algebra — cjohnson @ 7:18 pm Last time we defined the eigenvalues and eigenvectors of a matrix, but didn’t really discuss how to actually calculate the eigenvalues or […]

    Pingback by The Characteristic Polynomial « Mathematics Prelims — June 26, 2009 @ 7:18 pm | Reply

  4. Now I am a little disagree with this formulation
    “So a matrix is diagonalizable (similar to a diagonal matrix) if and only if its eigenvectors form a basis for the vector space”
    Since there are infinitely many eigenvectors, you need to tell which eigenvectors that form a basis or you can avoid that by saying that there are eigenvectors that form a basis.

    Thanks

    Comment by watchmath — June 28, 2009 @ 7:02 am | Reply


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