Linear algebra practice

Try a new question, then work through the fixed examples below. If you get stuck, check one step and retry.

Use $A=\begin{bmatrix}1&0\\2&1\\3&1\end{bmatrix}$ and $y=[2,5,8]^\mathsf T$ where specified. Try each question before opening its answer.

1. Read an entry

Given $A\in\mathbb R^{3\times2}$, find $A_{32}$ and explain its meaning.

Answer

$A_{32}=1$: record 3 has one practice set. The first index selects a row; the second selects a column.

2. Compare records

Find $x_3-x_1$.

Answer

$[3,1]^\mathsf T-[1,0]^\mathsf T=[2,1]^\mathsf T$. The third record has two more study hours and one more set.

3. Decode a weighted sum

Evaluate $\sum_{j=1}^{2}w_jx_j$ for $w=[-1,3]^\mathsf T$ and $x=[2,1]^\mathsf T$.

Answer

$(-1)(2)+(3)(1)=1$. The answer is a scalar, not a two-entry vector.

4. Check a product before calculating

If $B$ has shape $2\times4$, which of $AB$ and $BA$ are defined? Give each defined product’s shape.

Answer

$AB$ is defined and has shape $3\times4$. $BA$ is undefined because its inner dimensions are 4 and 3.

5. Find an exact solution

Find $w$ satisfying $Aw=[3,5,8]^\mathsf T$, or explain why none exists.

Answer

Row 1 gives $w_1=3$. Row 2 gives $6+w_2=5$, so $w_2=-1$. Row 3 then gives $9-1=8$. The solution is $[3,-1]^\mathsf T$.

6. Recognize an impossible target

Does $[2,5,8]^\mathsf T$ lie in the column space of $A$?

Answer

No. Every vector $Aw$ has its third entry equal to the sum of its first two; $8\ne2+5$. Equivalently, the first two equations require $w=[2,1]^\mathsf T$, which predicts 7 in the third row.

7. Distinguish two meanings of independence

The columns of $C$ are $a=[1,2,3]^\mathsf T$ and $2a$. Are they linearly independent? Is this a question about probabilistic independence?

Answer

They are linearly dependent because $2a-1(2a)=0$ uses coefficients that are not all zero. This is a relationship between fixed vectors, not a claim about independent random events.

8. Check perpendicularity

For $a=[1,2]^\mathsf T$ and $r=[-2/5,1/5]^\mathsf T$, compute $a^\mathsf Tr$.

Answer

$-2/5+2/5=0$, so the vectors are perpendicular in Euclidean geometry.

9. Distinguish a coefficient from a projection

Project $y_0=[2,5]^\mathsf T$ onto the span of $a=[1,2]^\mathsf T$. Report both the coefficient and the projected vector.

Answer

The coefficient is $t=(a^\mathsf Ty_0)/(a^\mathsf Ta)=12/5$. The projected vector is $ta=[12/5,24/5]^\mathsf T$. A scalar coefficient and a vector prediction are different answers.

10. Distinguish argmin from minimum

For $\min_w\lVert y-Aw\rVert^2$, report the minimum value and one minimizing weight vector.

Answer

The minimum value is $1/3$. The minimizing vector is $[7/3,2/3]^\mathsf T$. If the objective were MSE, dividing by three would change the minimum to $1/9$ but leave the weights unchanged.

11. Add an intercept

We want predictions $b+w_1h_i+w_2s_i$. Write the new design matrix and its shape, placing the intercept coefficient first.

Answer

The coefficient vector is $[b,w_1,w_2]^\mathsf T$, and the matrix is $\begin{bmatrix}1&1&0\\1&2&1\\1&3&1\end{bmatrix}$, of shape $3\times3$. The first column multiplies the same intercept into every record. Changing the feature set changes the fitting problem.

12. Identify the missing evidence

A model has training MSE $1/9$. What is its MSE on future records?

Answer

The training score does not determine it. We need new-data evidence, such as evaluation on an appropriate held-out set. Least squares solves the stated training objective, not the separate question of future performance.