Diagonalization of matrices

References: edX online course MIT 8.05 Week 6.

Suppose we have an operator {T} that has a matrix representation {T\left(\left\{ v\right\} \right)} in some basis {v}. In some cases (not all!) it is possible to transform to a different basis {u} in which {T\left(\left\{ u\right\} \right)} is a diagonal matrix. If the operator {A} transforms from the basis {v} to {u}, then we’ve seen that {T} transforms according to

\displaystyle  T\left(\left\{ u\right\} \right)=A^{-1}T\left(\left\{ v\right\} \right)A \ \ \ \ \ (1)

The diagonalization problem is therefore to find the matrix {A} (if it exists).

Assuming {A} does exist, we can look at the situation in two ways.

  1. The matrix representation of {T} is diagonal in the {u} basis.
  2. The operator {A^{-1}TA} is diagonal in the original {v} basis.

To prove (2), we start with the fact that {T} is diagonal in the {u} basis, so that (no implied sums in what follows):

\displaystyle   Tu_{i} \displaystyle  = \displaystyle  \lambda_{i}u_{i}\ \ \ \ \ (2)
\displaystyle  TAv_{i} \displaystyle  = \displaystyle  \lambda_{i}Av_{i}\ \ \ \ \ (3)
\displaystyle  A^{-1}TAv_{i} \displaystyle  = \displaystyle  \lambda_{i}A^{-1}Av_{i}\ \ \ \ \ (4)
\displaystyle  \displaystyle  = \displaystyle  \lambda_{i}v_{i} \ \ \ \ \ (5)

From the last line, we see that {v_{i}} is an eigenvector of the operator {A^{-1}TA} with the same eigenvalue {\lambda_{i}} as the eigenvector {u_{i}} of the operator {T}.

Thus we can write, in the original basis {v}, the equation

\displaystyle  D_{T}\equiv A^{-1}TA \ \ \ \ \ (6)

where {D_{T}} is the diagonalized version of the matrix {T}. This is known as a similarity transformation.

All this is fine, but we still haven’t seen how to find {A}. It turns out that the columns of {A} are the eigenvectors of {T}. We can see this from the following argument.

In the basis {v}, each {v_{i}} has the form of a column vector with a 1 in the {i}th position and zeroes everywhere else. Thus the transformation to the {u} basis can be written as

\displaystyle  u_{k}=Av_{k}=\left[\begin{array}{ccc} A_{11} & \ldots & A_{1n}\\ \vdots & \ddots & \vdots\\ A_{n1} & \ldots & A_{nn} \end{array}\right]\left[\begin{array}{c} \vdots\\ 1\\ \vdots \end{array}\right]=\left[\begin{array}{c} A_{1k}\\ \vdots\\ A_{nk} \end{array}\right] \ \ \ \ \ (7)

The 1 is in the {k}th position in the column vector representing {v_{k}} and picks out the elements in column {k} of {A} in the product {Av_{k}}. Since {u_{k}} is the {k}th eigenvector of {T}, the columns of {A} are the eigenvectors of {T}.

Example Suppose we have

\displaystyle  T=\left[\begin{array}{cc} 1 & 2\\ 3 & 1 \end{array}\right] \ \ \ \ \ (8)

The eigenvalues are found in the usual way, from the characteristic determinant, which is

\displaystyle   \left(1-\lambda\right)^{2}-6 \displaystyle  = \displaystyle  0\ \ \ \ \ (9)
\displaystyle  \lambda \displaystyle  = \displaystyle  1\pm\sqrt{6} \ \ \ \ \ (10)

We can find the eigenvectors by solving {Tu_{i}=\lambda_{i}u_{i}} for each eigenvalue, where {u_{i}} is a 2-element column vector. (I won’t go through this, since it’s just algebra.) Placing the eigenvectors as the columns in a matrix {A}, we have

\displaystyle  A=\left[\begin{array}{cc} \frac{\sqrt{6}}{3} & -\frac{\sqrt{6}}{3}\\ 1 & 1 \end{array}\right] \ \ \ \ \ (11)

The inverse of a {2\times2} matrix {A=\left[\begin{array}{cc} a & b\\ c & d \end{array}\right]} is

\displaystyle  A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc} d & -b\\ -c & a \end{array}\right] \ \ \ \ \ (12)

So for our matrix, we have

\displaystyle  A^{-1}=\left[\begin{array}{cc} \frac{\sqrt{6}}{4} & \frac{1}{2}\\ -\frac{\sqrt{6}}{4} & \frac{1}{2} \end{array}\right] \ \ \ \ \ (13)

Doing the matrix products (just a lot of arithmetic) we get

\displaystyle  A^{-1}TA=\left[\begin{array}{cc} 1+\sqrt{6} & 0\\ 0 & 1-\sqrt{6} \end{array}\right] \ \ \ \ \ (14)

The resulting matrix is diagonal, and the diagonal entries are the eigenvalues of {T}. It’s worth noticing that the traces of {A^{-1}TA} and {T} are the same (both = 2), as are the determinants (both {=-5}).

3 thoughts on “Diagonalization of matrices

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