2.9 Exercises

20 points · 4 extra credit

Dimension and Rank

1. 1. (3 pts) For

   $A = \begin{bmatrix} 1 & 3 & 2 & 4 \\ 2 & 6 & 5 & 7 \\ 1 & 3 & 1 & 5 \end{bmatrix},$

   find $\operatorname{rank} A$, $\dim \operatorname{Col} A$, and $\dim \operatorname{Nul} A$, and verify the Rank Theorem $\operatorname{rank} A + \dim \operatorname{Nul} A = n$.

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2. 2. (3 pts) A matrix $A$ is $5 \times 9$ and $\dim \operatorname{Nul} A = 4$. Find $\operatorname{rank} A$ and $\dim \operatorname{Col} A$. Do the columns of $A$ span $\mathbb{R}^5$?

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3. 3. (2 pts) A matrix $A$ is $7 \times 5$ with $\operatorname{rank} A = 5$. Are the columns of $A$ linearly independent? Is $A\mathbf{x} = \mathbf{b}$ consistent for every $\mathbf{b} \in \mathbb{R}^7$? Explain both answers.

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4. 4. (3 pts) State whether each is true or false, and justify.

   (a) If $A$ is $4 \times 6$, then $\operatorname{rank} A \le 4$.

   (b) If a $5 \times 5$ matrix has rank $5$, then it is invertible.

   (c) For any matrix $A$, $\dim \operatorname{Col} A = \dim \operatorname{Nul} A$.

   (d) If $\operatorname{rank} A$ equals the number of columns of $A$, then $A\mathbf{x} = \mathbf{0}$ has only the trivial solution.

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5. 5. (3 pts) The set $\mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix} \right\}$ is a basis for $\mathbb{R}^2$. Find the coordinate vector $[\mathbf{x}]_{\mathcal{B}}$ of $\mathbf{x} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$.

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6. 6. (3 pts) Explain why, for any $m \times n$ matrix $A$, the number of pivot columns plus the number of non-pivot columns equals $n$, and use this to explain the Rank Theorem $\operatorname{rank} A + \dim \operatorname{Nul} A = n$.

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7. 7. (3 pts) Let $A$ be a $5 \times 5$ matrix.

   (a) If $\operatorname{rank} A = 5$, is $A$ invertible? Cite the relevant statement of the Invertible Matrix Theorem.

   (b) If $\operatorname{rank} A = 4$, what is $\dim \operatorname{Nul} A$, and is $A$ invertible?

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8. 8. (4 pts ••) Let $A$ be $m \times n$ and $B$ be $n \times p$. Prove that

   $\operatorname{rank}(AB) \le \min\{\operatorname{rank} A,\ \operatorname{rank} B\}.$

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