Linear Algebra and Its Applications, Exercise 2.5.5

Exercise 2.5.5. Given the incidence matrix A from exercise 2.5.1 and the diagonal matrix

C = \begin{bmatrix} c_1&0&0 \\ 0&c_2&0 \\ 0&0&c_3 \end{bmatrix}

compute A^TCA and show that the 2 by 2 matrix resulting from removing the third row and third column is invertible..

Answer: From exercise 2.5.1 we have the incidence matrix

A = \begin{bmatrix} 1&-1&0 \\ 0&1&-1 \\ 1&0&-1 \end{bmatrix}

so that

A^TCA = \begin{bmatrix} 1&0&1 \\ -1&1&0 \\ 0&-1&-1 \end{bmatrix} \begin{bmatrix} c_1&0&0 \\ 0&c_2&0 \\ 0&0&c_3 \end{bmatrix} \begin{bmatrix} 1&-1&0 \\ 0&1&-1 \\ 1&0&-1 \end{bmatrix}

= \begin{bmatrix} c_1&0&c_3 \\ -c_1&c_2&0 \\ 0&-c_2&-c_3 \end{bmatrix} \begin{bmatrix} 1&-1&0 \\ 0&1&-1 \\ 1&0&-1 \end{bmatrix}

= \begin{bmatrix} c_1+c_3&-c_1&-c_3 \\ -c_1&c_1+c_2&-c_2 \\ -c_3&-c_2&c_2+c_3 \end{bmatrix}

Note that the third row of this matrix is equal to -1 times the sum of the first and second rows, so the rows are linearly dependent and the matrix is singular.

If we remove the third row and third column of A^TCA we obtain the following matrix:

\begin{bmatrix} c_1+c_3&-c_1 \\ -c_1&c_1+c_2 \end{bmatrix}

This matrix is nonsingular (except for certain values of c_1, c_2 and c_3 as discussed below); its inverse is

1/((c_1+c_3)(c_1+c_2) - (-c_1)(-c_1)) \begin{bmatrix} c_1+c_2&-(-c_1) \\ -(-c_1)&c_1+c_3 \end{bmatrix}

= 1/(c_1^2+c_1c_2+c_3c_1+c_3c_2-c_1^2) \begin{bmatrix} c_1+c_2&c_1 \\ c_1&c_1+c_3 \end{bmatrix}

1/(c_1c_2+c_1c_3+c_2c_3) \begin{bmatrix} c_1+c_2&c_1 \\ c_1&c_1+c_3 \end{bmatrix}

Note that if c_1c_2+c_1c_3+c_2c_3 = 0 then the 2 by 2 matrix is singular and has no inverse. This would be true, for example, if c_1 = c_2 = 2 and c_3 = -1 so that the 2 by 2 matrix derived from A^TCA is

\begin{bmatrix} 1&-2 \\ -2&4 \end{bmatrix}

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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