## 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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