Corrected exercises: basis, dimension and the fundamental subspaces (MATH 133)

The coefficient matrix is the single row $(1 \;\; 2 \;\; -1)$, already reduced, with its pivot on $x$. The variables $y$ and $z$ are free: $x = -2y + z$, so $(x, y, z) = y(-2, 1, 0) + z(1, 0, 1)$. Every vector of $P$ is a combination of $(-2, 1, 0)$ and $(1, 0, 1)$, and these two are independent because each has a $1$ in the free position where the other has a $0$ (look at the $y$ and $z$ entries). So $\{(-2, 1, 0), (1, 0, 1)\}$ is a basis and $\dim P = 2$: three unknowns, one independent equation, two free variables. This is exactly the null space of the $1 \times 3$ matrix $(1 \;\; 2 \;\; -1)$, and rank plus nullity says $1 + 2 = 3$.

Membership is one substitution each: $0 + 2(1) - 2 = 0$ and $3 + 2(-1) - 1 = 0$. The two vectors are not multiples of one another (the first entry of $u$ is $0$, that of $w$ is not, while their second entries are both non-zero), so they are independent. Now the dimension does the rest: $P$ has dimension $2$, and in a subspace of dimension $m$ any $m$ independent vectors automatically span. So $\{u, w\}$ is a basis of $P$, with no spanning computation at all. Checking both conditions is not wrong, it is wasted time; forgetting to say WHY one check suffices (the dimension is $2$) is what costs the mark.

Solve $au + bw = v$: $(3b, \; a - b, \; 2a + b) = (3, 1, 5)$. The first entry gives $b = 1$, the second $a = 2$, and the third checks: $2(2) + 1 = 5$. So $v = 2u + w$ and $[v]_{\{u,w\}} = (2, 1)$. In the basis of a), the coordinates are simply the free variables: $y = 1$, $z = 5$, and indeed $1 \cdot (-2, 1, 0) + 5 \cdot (1, 0, 1) = (3, 1, 5)$. So $[v] = (1, 5)$ there. Same vector, two different coordinate pairs: coordinates belong to a basis, not to a vector. And note both pairs have TWO entries, not three, because $v$ is being described inside a plane.

No, and no computation is needed to say so: $P$ has dimension $2$, and any three vectors in a subspace of dimension $2$ are dependent. To exhibit the dependency, solve $au + bw = (-2, 1, 0)$: $3b = -2$ gives $b = -\frac{2}{3}$, and $2a + b = 0$ gives $a = \frac{1}{3}$; the middle entry checks, $\frac{1}{3} + \frac{2}{3} = 1$. So $(-2, 1, 0) = \frac{1}{3}u - \frac{2}{3}w$, that is $3(-2, 1, 0) = u - 2w$.

Any vector outside $P$ will do. Take $e_1 = (1, 0, 0)$: $1 + 0 - 0 = 1 \neq 0$, so $e_1 \notin P$. Then $\{u, w, e_1\}$ is independent (a vector outside the span of an independent pair cannot be a combination of it), and three independent vectors in $\mathbb{R}^3$, of dimension $3$, form a basis. The one-line justification the marker wants is precisely that last clause: three independent vectors in a space of dimension three.

The matrix has rows $(1, 0, 2, 1, 1)$, $(2, 1, 3, 3, 0)$, $(-1, 1, -3, 0, 0)$, $(0, 1, -1, 1, 2)$. First column: $R_2 \to R_2 - 2R_1$ gives $(0, 1, -1, 1, -2)$ and $R_3 \to R_3 + R_1$ gives $(0, 1, -1, 1, 1)$. Second column: $R_3 \to R_3 - R_2$ gives $(0, 0, 0, 0, 3)$ and $R_4 \to R_4 - R_2$ gives $(0, 0, 0, 0, 4)$. Then $R_3 \to \frac{1}{3}R_3$ gives $(0, 0, 0, 0, 1)$ and $R_4 \to R_4 - 4R_3$ kills the last row. Clearing above the third pivot: $R_2 \to R_2 + 2R_3$ gives $(0, 1, -1, 1, 0)$ and $R_1 \to R_1 - R_3$ gives $(1, 0, 2, 1, 0)$. The reduced form is $\begin{pmatrix} 1 & 0 & 2 & 1 & 0 \\ 0 & 1 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$, with pivots in columns $1$, $2$ and $5$.

Keep the ORIGINAL vectors in the pivot columns: $\{v_1, v_2, v_5\}$ is a basis of $W$, and $\dim W = 3$. Why this works: row operations do not change the linear relations between the columns. A relation $\sum c_i v_i = 0$ holds for the columns of the original matrix exactly when it holds for the columns of $R$, because both say that $(c_1, \dots, c_5)$ solves the same homogeneous system. On $R$, the pivot columns are visibly independent and every other column is visibly a combination of them, so the same is true of $v_1, v_2, v_5$.

Column $3$ of $R$ is $(2, -1, 0, 0)$, which is $2 \times$(column $1$) $- 1 \times$(column $2$). The same relation holds among the original columns: $v_3 = 2v_1 - v_2$. Check: $2(1, 2, -1, 0) - (0, 1, 1, 1) = (2, 3, -3, -1)$. Column $4$ of $R$ is $(1, 1, 0, 0)$, so $v_4 = v_1 + v_2 = (1, 3, 0, 1)$. The non-pivot columns of $R$ are not decorations: they are the recipe expressing each discarded vector in the kept ones.

Substitute each into $-6y_1 + y_2 - 4y_3 + 3y_4$: $e_1 = (1, 0, 0, 0)$ gives $-6$, $e_2 = (0, 1, 0, 0)$ gives $1$, $e_3 = (0, 0, 1, 0)$ gives $-4$. None gives $0$, so NOT ONE of the student's three vectors belongs to $W$. The equation is right, too: $v_1$ gives $-6 + 2 + 4 + 0 = 0$, $v_2$ gives $0 + 1 - 4 + 3 = 0$ and $v_5$ gives $-6 + 0 + 0 + 6 = 0$. The student has the right NUMBER, three, because row operations preserve the dimension, and the wrong vectors, because they do not preserve the column space. That answer scores zero on the basis and full marks on the dimension.

In the order $v_5, v_4, v_3, v_2, v_1$, the method keeps $v_5$; then $v_4$, which is not a multiple of $v_5$; then asks whether $v_3 = (2, 3, -3, -1)$ is a combination $a v_5 + b v_4 = (a + b, 3b, 0, 2a + b)$. The third entry of any such combination is $0$ and that of $v_3$ is $-3$, so $v_3$ is new and is kept. Three pivots are already reached, and $\dim W = 3$, so $v_2$ and $v_1$ are discarded. The extracted basis is $\{v_5, v_4, v_3\}$: a different basis, the same size. Which vectors survive depends on the order; how many survive does not, and that invariant is the dimension.

First column: $R_2 \to R_2 - 2R_1$ gives $(0, 0, -1, 0, -2)$ and $R_3 \to R_3 + R_1$ gives $(0, 0, 1, 1, 3)$; $R_4 = (0, 0, 1, 1, 3)$ already starts with zeros. Column $2$ has nothing below row $1$, so the next pivot is in column $3$: $R_2 \to -R_2$ gives $(0, 0, 1, 0, 2)$, then $R_3 \to R_3 - R_2$ and $R_4 \to R_4 - R_2$ both give $(0, 0, 0, 1, 1)$. Then $R_4 \to R_4 - R_3$ gives a zero row. Finally $R_1 \to R_1 - R_2$ clears above the second pivot: $(1, 2, 0, 0, -1)$. So $R = \begin{pmatrix} 1 & 2 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}$, with pivots in columns $1$, $3$ and $4$: rank $3$.

Solve $Rx = 0$. The free variables are $x_2 = s$ and $x_5 = t$; the rows read $x_1 = -2s + t$, $x_3 = -2t$, $x_4 = -t$. So $x = s(-2, 1, 0, 0, 0) + t(1, 0, -2, -1, 1)$, and $\{(-2, 1, 0, 0, 0), (1, 0, -2, -1, 1)\}$ is a basis of $\text{null}(A)$, one vector per free variable, independent because each carries a $1$ in its own free slot and a $0$ in the other. Nullity $2$. Check the second on the ORIGINAL $A$, row by row: $1 + 0 - 2 + 0 + 1 = 0$, $2 + 0 - 2 + 0 + 0 = 0$, $-1 + 0 + 0 - 1 + 2 = 0$, $0 + 0 - 2 - 1 + 3 = 0$. Here using $R$ is legitimate: $Ax = 0$ and $Rx = 0$ have the same solutions.

The pivots sit in columns $1$, $3$ and $4$, so the basis of $\text{col}(A)$ is made of columns $1$, $3$ and $4$ OF $A$: $\{(1, 2, -1, 0), (1, 1, 0, 1), (0, 0, 1, 1)\}$, and $\dim \text{col}(A) = 3$. The reduced form told us which positions to keep and how many; the vectors must be vectors of $\text{col}(A)$, and only the columns of $A$ are guaranteed to be. The discarded columns are combinations of the kept ones with the coefficients read on $R$: column $2$ of $A$ is $2 \times$ column $1$, and column $5$ is $-$column $1 + 2 \times$column $3 +$column $4$, that is $(-1 + 2 + 0, -2 + 2 + 0, 1 + 0 + 1, 0 + 2 + 1) = (1, 0, 2, 3)$, which is indeed column $5$.

Row operations replace rows by combinations of rows and are reversible, so they do not change the row space. The non-zero rows of $R$ are independent (each has a leading $1$ where the others have $0$), so they form a basis: $\{(1, 2, 0, 0, -1), (0, 0, 1, 0, 2), (0, 0, 0, 1, 1)\}$, dimension $3$. The same number $3$ as for $\text{col}(A)$, although the two subspaces do not even live in the same space: $\text{col}(A) \subseteq \mathbb{R}^4$, $\text{row}(A) \subseteq \mathbb{R}^5$. Rank plus nullity: $3 + 2 = 5$, the number of COLUMNS of $A$, as it must be.

Try to write $e_1 = a(1, 2, -1, 0) + b(1, 1, 0, 1) + c(0, 0, 1, 1)$. Entry by entry: $a + b = 1$, $2a + b = 0$, $-a + c = 0$, $b + c = 0$. Subtracting the first from the second gives $a = -1$, so $b = 2$, then $c = a = -1$, and the last equation demands $2 - 1 = 0$: impossible. So $e_1 \notin \text{col}(A)$. Every column of $A$ satisfies $y_4 = y_1 + y_3$ (check: $0 = 1 - 1$, $0 = 2 - 2$, $1 = 1 + 0$, $1 = 0 + 1$, $3 = 1 + 2$), and $e_1$ does not. Quoting the pivot columns of $R$ as a basis of $\text{col}(A)$ would therefore give a family that is not even inside the subspace: zero on part c), however clean the reduction.

Read $Rx = 0$: the pivots are in columns $1$ and $3$, so $x_2 = s$ and $x_4 = t$ are free. The rows give $x_1 = -2s - t$ and $x_3 = 2t$. Hence $x = s(-2, 1, 0, 0) + t(-1, 0, 2, 1)$ and $\{(-2, 1, 0, 0), (-1, 0, 2, 1)\}$ is a basis of $\text{null}(A)$, nullity $2$. Check the second vector on $A$: $-1 + 0 - 2 + 3 = 0$, $-2 + 0 - 4 + 6 = 0$, $-1 + 0 + 0 + 1 = 0$.

Columns $1$ and $3$ of $A$, namely $(1, 2, 1)$ and $(-1, -2, 0)$, dimension $2$. Not columns $1$ and $3$ of $R$: those are $(1, 0, 0)$ and $(0, 1, 0)$, and every column of $A$ has its second entry equal to twice its first ($2 = 2 \cdot 1$, $4 = 2 \cdot 2$, $-2 = 2 \cdot (-1)$, $6 = 2 \cdot 3$), because row $2$ of $A$ is twice row $1$. So $\text{col}(A)$ lies in the plane $y_2 = 2y_1$, and $(0, 1, 0)$ violates it. Same trap as always, and the figure is there to make it visible: the boxes mark positions in $R$, the vectors are taken from $A$.

The student assumes that row $i$ of $R$ comes from row $i$ of $A$. It does not: the reduction did $R_2 \to R_2 - 2R_1$, which made row $2$ zero at once, then $R_3 \to R_3 - R_1$ and a SWAP of rows $2$ and $3$, so the zero row of $R$ is the old row $2$. And rows $1$ and $2$ of $A$, $(1, 2, -1, 3)$ and $(2, 4, -2, 6)$, are proportional: they span a line, not a space of dimension $2$. The safe basis is the non-zero rows of $R$: $\{(1, 2, 0, 1), (0, 0, 1, -2)\}$. A basis made of rows of $A$ also exists: $\text{row}(A)$ has dimension $2$, and rows $1$ and $3$, $(1, 2, -1, 3)$ and $(1, 2, 0, 1)$, are two independent vectors of it (third entries $-1$ and $0$ with equal first entries), so they form a basis by the dimension argument.

$\text{rank}(A^T) = \text{rank}(A) = 2$, because the columns of $A^T$ are the rows of $A$: $\dim \text{col}(A^T) = \dim \text{row}(A) = \dim \text{col}(A)$. And $\text{col}(A^T)$ is literally $\text{row}(A)$ with the vectors written vertically, so a basis is $(1, 2, 0, 1)$ and $(0, 0, 1, -2)$ as column vectors of $\mathbb{R}^4$. No reduction of the $4 \times 3$ matrix $A^T$ is needed, and doing one would cost time without adding a single fact.

$A^T$ is $4 \times 3$, so rank plus nullity counts its $3$ columns: $\dim \text{null}(A^T) = 3 - 2 = 1$. A vector $y = (y_1, y_2, y_3)$ with $A^T y = 0$ is exactly a relation $y_1 \text{row}_1 + y_2 \text{row}_2 + y_3 \text{row}_3 = 0$ among the rows of $A$. The relation is in plain sight: $\text{row}_2 = 2\,\text{row}_1$, so $y = (-2, 1, 0)$. Since the dimension is $1$, this single vector is even a basis. It is also the coefficient vector of the equation $-2y_1 + y_2 = 0$ that carves $\text{col}(A)$ out of $\mathbb{R}^3$, the same fact seen from the other side.