Linear Algebra and Its Applications, Exercise 2.5.4

Exercise 2.5.4. Given the incidence matrix $A$ from exercise 2.5.1 show that $A^TA$ is symmetric and singular, and determine its nullspace. Show that the matrix obtained by removing the last row and column of $A^TA$ is nonsingular.

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

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

$= \begin{bmatrix} 2&-1&-1 \\ -1&2&-1 \\ -1&-1&2 \end{bmatrix}$

and $A^TA$ is symmetric.

To see whether $A^TA$ is singular we perform Gaussian elimination on it. We start by multplying the first row by $-\frac{1}{2}$ and subtracting it from the second and third rows:

$= \begin{bmatrix} 2&-1&-1 \\ -1&2&-1 \\ -1&-1&2 \end{bmatrix} \Rightarrow \begin{bmatrix} 2&-1&-1 \\ 0&\frac{3}{2}&-\frac{3}{2} \\ 0&-\frac{3}{2}&\frac{3}{2} \end{bmatrix}$

We then multiply the second row by -1 and subtract it from the third row:

$\begin{bmatrix} 2&-1&-1 \\ 0&\frac{3}{2}&-\frac{3}{2} \\ 0&-\frac{3}{2}&\frac{3}{2} \end{bmatrix} \Rightarrow \begin{bmatrix} 2&-1&-1 \\ 0&\frac{3}{2}&-\frac{3}{2} \\ 0&0&0 \end{bmatrix}$

The matrix is now in echelon form. Since it has three columns and only two pivots the matrix is singular (with rank $r = 2$ ).

In solving $(A^TA)x = 0$ we have $x_1$ and $x_2$ as basic variables and $x_3$ as a free variable. Using the echelon matrix above we have

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr} 2x_1&-&x_2&-&x_3&=&0 \\ &&\frac{3}{2}x_2&-&\frac{3}{2}x_3&=&0 \end{array}$

From the second equation we have $x_2 = x_3$ and substituting into the first equation we have $x_1 = x_2 = x_3$ . The nullspace therefore consists of all vectors of the form $\begin{bmatrix} c&c&c \end{bmatrix}^T$ . It has dimension $n - r = 3 - 2 = 1$ .

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

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

This matrix is nonsingular; its inverse is

$1/(2 \cdot 2 - (-1) \cdot (-1)) \begin{bmatrix} 2&-(-1) \\ -(-1)&2 \end{bmatrix}$

$= 1/(4-1) \begin{bmatrix} 2&1 \\ 1&2 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 2&1 \\ 1&2 \end{bmatrix} = \begin{bmatrix} \frac{2}{3}&\frac{1}{3} \\ \frac{1}{3}&\frac{2}{3} \end{bmatrix}$

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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Linear Algebra and Its Applications, Exercise 2.5.4

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