Solutions to Midterm #2

Questions on grading are welcome, even for one point. Please write any questions on the front of the exam and hand it back to me by the end of week 9.

Again, no letter grades are actually recorded, but estimated ranges would be C- to C 18-26, B- to B+ 27-37, A- to A 38-50.

All problems were from homework or ``proofs to know''. References:

1. Hw 6, p. 26, Ex. 1; see p. Y3.

2. (a) Hw 5, p. 73 Ex. 8; (b) Hw 7 p. 95, Ex. 2.

3. (a) Hw 6, Q-7; (b) from proofs to know.

4. From proofs to know.

5. (a) Hw 6, R-2 (see p. Y9); (b) Hw 5, O-3; (c) Hw 4, J-3 and Hw 6, Q-1; (d) Hw 6, p. 21, Ex. 7 and Hw 7, V-9.

Solutions.

1. Method #1: Find the elementary matrices which, when multiplied on the left, change $A$ to RREF. Then multiply the elementary matrices together to get $P$. [Details omitted]

Method #2: Row-reduce $[A\vert I] \rightsquigarrow [PA\vert P]$, as follows:

$\left[\begin{array}{rrrrrrrr}1&2&1&5&9&1&0&0\\ 1&2&2&7&14&0&1&0\\ 1&2&1&5&9&... ...&0&0\\ 0&0&1&2&5&-1&1&0\\ 0&0&0&0&0&-1&0&1\end{array}\right] \rightsquigarrow$

$\left[\begin{array}{rrrrrrrr}1&2&0&3&4&2&-1&0\\ 0&0&1&2&5&-1&1&0\\ 0&0&0&0&0&-1&0&1\end{array}\right]$, so the answer is $P = \left[\begin{array}{rrr}2&-1&0\\ -1&1&0\\ -1&0&1\end{array}\right]$.

Check: $PA = \left[\begin{array}{rrr}2&-1&0\\ -1&1&0\\ -1&0&1\end{array}\right]\left... ...\left[\begin{array}{rrrrr}1&2&0&3&4\\ 0&0&1&2&5\\ 0&0&0&0&0\end{array}\right]$,
which is in RREF.

2. (a) Let $T = \tau _ A$ with $A = \left[\begin{array}{rrr}2&8&0\\ 3&1&0\\ 7&4&0\end{array}\right]$.

Reason this is valid: The range of $\tau _ A$ is the column space of $A$, which is the span of the first two columns of $A$ since the third column is 0.

(b) $\tau _ A (v _ 1) = \left[\begin{array}{rr}1&1\\ 0&1\end{array}\right]\left[\b... ...{r}1\\ 1\end{array}\right] + s \left[\begin{array}{r}1\\ -1\end{array}\right]$. This gives $r + s = 2$, $r - s = 1$; solving we get $2r = 3$, $r = \frac 32$, $s = \frac 12$.

$\tau _ A (v _ 2) = \left[\begin{array}{rr}1&1\\ 0&1\end{array}\right]\left[\b... ...{r}1\\ 1\end{array}\right] + u \left[\begin{array}{r}1\\ -1\end{array}\right]$. This gives $t + u = 0$, $t - u = -1$; solving we get $2t = -1$, $t = -\frac 12$, $u = \frac 12$.

Putting these results together, we see that the matrix of $T$ with respect to the basis $v _ 1, v _ 2$ is $M = \left[\begin{array}{rr} \frac 32& -\frac 12\\ \frac 12& \frac 12\end{array}\right]$.

3. (a) $v = 5w$, so $w = \frac 15 v$. But which element is $\frac 15$ (the multiplicative inverse of 5) in $\mathbb{Z}_ {31}$? Using the Euclidean algorithm idea, we have

$\left[\begin{array}{rrr}31&1&0\\ 5&0&1\end{array}\right] \rightsquigarrow \le... ...t] \rightsquigarrow \left[\begin{array}{rrr}1&1&-6\\ 0&-5&31\end{array}\right]$. Expressing $(1,1,-6)$ as a linear combination of $(31,1,0)$ and $(5,0,1)$ we get $(1,1,6) = 1 \cdot (31,1,0) + (-6) \cdot (5,0,1)$, so $1 = 1 \cdot 31 + (-6) \cdot 5$. Working modulo 31, this becomes $1 = 31 (-6)$ in $\mathbb{Z}_ {31}$. Since $-6 = 25$ in $\mathbb{Z}_ {31}$, the answer is that $w = 25v$.

(b) We use the Lemma that a list of vectors $v _ 1,\dots, v _ n$ is linearly dependent $\Leftrightarrow$ some vector is in the span of the preceding vectors.

Consider $w _ 2, w _ 1, v _ 1, v _ 2, v _ 3$. Since the original list of vectors already spanned $V$, $w _ 2$ is in that span, so this list is linearly dependent. Then by the Lemma, some vector in the list is in the span of the preceding vectors. That vector can't be $w _ 1$ or $w _ 2$, since they are in a linearly independent set of vectors and so are linearly independent. Therefore it must be some $v _ i$. Then that $v _ i$ can be removed without changing the span, which is therefore still $V$.

4. (a) Since $w _ 1, w _ 2$ are in the range of $T$, there are $v _ 1$, $v _ 2$ with $T(v _ i) = w _ i$ for $i = 1,2$. The proposed basis is $u _ 1, u _ 2, u _ 3, v _ 1, v _ 2$.

(b) To show that the vectors in (a) are linearly independent, we suppose that a linear combination of them is 0 and try to show that the coefficients are all zero. So suppose that for scalars $r _ 1, r _ 2, r _ 3, s _ 1, s _ 2$ we have

(*) $r _ 1 u _ 1 + r _ 2 u _ 2 + r _ 3 u _ 3 + s _ 1 w _ 1 + s _ 2 w _ 2 =$   0.

Applying $T$ and using the fact that $T$ is linear, we get

0$+ s _ 1 w _ 1 + s _ 2 w _ 2 =$   0 in $W$. Since $w _ 1, w _ 2$ are a basis of Range$(T)$, they are linearly independent, so $s _ 1 = s _ 2 = 0$. Then (*) becomes
0$+ r _ 1 u _ 1 + r _ 2 u _ 2 + r _ 3 u _ 3 =$   0. Since $u _ 1, u _ 2, u _ 3$ are a basis for the null space of $T$, they are linearly independent, so $s _ 1 = s _ 2 = 0$.

Therefore in (*) all coefficients must be 0, and we have proved the linear independence of the vectors from (a).

5. (a) We know $p(D) = \left[\begin{array}{ccc}p(1)&0&0\\ 0&p(0)&0\\ 0&0&p(-1)\end{array}\right]$, so we want $p(1) = 0$, $p(0) = 0$, $p(-1)= 0$. In other words, $p(x)$ has roots $1,0,-1$. Then an answer is $p(x) = (x - 1)(x - 0)(x + 1) = x^3 - x$.

(b) We know that (i) $T($subspace$)$ is a subspace and (ii) $T^{-1}($subspace$)$ is a subspace. Putting these together, we see that $T^{-1} (T(U))$ is a subspace. It's a subspace of $V$ since we went right and then left.

(Everyone needs more work on what $T^{-1}$ means. If $T$ is invertible then $T^{-1}$ is actually another linear transformation. But whether $T$ is invertible or not, if $S$ is a subspace of $W$, $T^{-1}(S)$ means the set of all $v \in V$ with $T(v) \in S$, and it's a subspace. This is an application of the notation $\inf f(S)$ for subsets.)

(c) This says that for any nonzero vector $v \in \mathbb{R}^ 2$ there is a $w \in \mathbb{R}_ 2$ so that $v w ^ t$ is nonsingular, i.e., has rank 2. This can't be true, since we know $v w ^ t$ has rank 1. (Recall from homework that $v w ^ t$ has rank 1 since the rows are all proportional to each other.)

(d) If $A = \left[\begin{array}{rr}a&b\\ c&d\end{array}\right]$ we know that $A^{-1} = {\frac{\displaystyle 1}{\displaystyle \Delta}} \left[\begin{array}{rr}d&-b\\ -c&a\end{array}\right]$, where $\Delta =$   det$\; A$. Since we are told $\Delta = 1$, the entries of $A^{-1}$ are all plus or minus entries of $A$, so are integers. Therefore the answer is ``yes''. (To do this problem, it's not necessary to remember the details of the entries of $A^{-1}$, just the general idea.)