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The pivot columns in the row echelon form are columns 1 and 2. Take the corresponding columns of the original matrix:For a direct check, call these columns $c_1,c_2$. The other columns satisfy $c_3=c_1-c_2$ and $c_4=c_1+c_2$, so they are in the same linear span. Also $c_1,c_2$ are linearly independent: their fourth entries force the coefficient of $c_2$ to vanish, then their first entries force the other coefficient to vanish. Elementary row operations preserve dependence relations among columns, but generally change the column space itself, which is why the original columns must be used.

For a real $m\times n$ matrix $A$, the row space is the linear span of its rows in ${\mathbb{R}}^n$, and the column space is the linear span of its columns in ${\mathbb{R}}^m$. Its null space consists of the input vectors sent to zero:The rank of a matrix isThe equality of row and column dimensions is the row-rank/column-rank theorem. Each displayed space is a linear subspace of the indicated ambient vector space.

Write the columns of $A$ as $a_1,\dots ,a_n$ and the coordinates of the vector $x$ as $x_1,\dots ,x_n$. By the definition of matrix multiplication,Indeed, its $i$th coordinate is ${\sum }_{j=1}^n{A}_{ij}{x}_j$, exactly the $i$th coordinate of this linear combination. Since the column space contains every linear combination of the columns, the claim holds for every vector $x$.

The pivot columns of the reduced row echelon form are columns 1 and 3. Take those columns of the original matrix, since elementary row operations need not preserve the original column space:For an explicit spanning check, column 2 is $-2a_1$, column 4 is $-a_1+2a_3$, and column 5 is $3a_1-2a_3$. The first two coordinates of the displayed columns form a $2\times 2$ matrix of determinant $-5\ne 0$, proving their linear independence. They are therefore a basis of the column space.

The right-hand vector is in the column space, becauseConsequently the linear system is consistent. A concrete solution isAs a compatibility check, the original rows satisfy $r_1+11r_2-5r_3=0$, and the right-hand entries satisfy the same relation $1+11\cdot 4-5\cdot 9=0$. There are infinitely many solutions, obtained by adding any vector in the null space to the displayed solution.

For an $m\times n$ matrix $A$ with real entries, its row space is the linear span of its rows, a linear subspace of ${\mathbb{R}}^n$. Its column space is the linear span of its columns, a linear subspace of ${\mathbb{R}}^m$ and the image of a linear map $x\mapsto Ax$. Its null space isa linear subspace of ${\mathbb{R}}^n$. Finally,Thus the rank of a matrix counts the number of linearly independent columns, equivalently rows. The nullity is $\dim \ker A=n-\mathrm{rank}A$ by the rank-nullity theorem.

The pivot columns in the row echelon form are columns 1 and 3. Use those columns of the original matrix for a basis of the original column space:For a direct spanning check, column 2 is twice column 1 and column 4 is $(17/5)$ times column 1 minus $(3/5)$ times column 3. The displayed columns are linearly independent, since the first two coordinates of $\alpha c_1+\beta c_3=0$ give $\alpha -\beta =0$ and $\alpha +4\beta =0$, so $\alpha =\beta =0$.

For a real $m\times n$ matrix $A$, the row space is the linear span of its rows in ${\mathbb{R}}^n$. The column space is the linear span of its columns in ${\mathbb{R}}^m$, equivalently $\{A\mathbf{x}:\mathbf{x}\in {\mathbb{R}}^n\}$. The null space isThe matrix rank is the dimension of the column space, which equals the dimension of the row space. In particular,The last equality is the rank-nullity theorem.

Let $c_1,\dots ,c_4$ be the original columns. The first three are linearly independent: restricting them to rows 2, 3 and 4 gives the triangular matrixwhose determinant is one. Also $c_4=c_1-c_2+c_3$, so a basis for the column space isThese are columns of the original matrix. Taking columns of its reduced row echelon form would generally give the wrong column space.

Denote the columns by $c_1,\dots ,c_4$. The first two are linearly independent, because the top two coordinates have determinant $3\cdot 2-(-1)\cdot 1=7$. Direct calculation givesThus a column-space basis is $\{(3,1,9{)}^T,(-1,2,4{)}^T\}$, andFor the last description, the third row equals twice the first row plus three times the second, so every column lies in that plane. Since the two independent columns already span a two-dimensional vector subspace, they fill the plane.

The target vector satisfies $5=2(-2)+3(3)$, so it is in the column space. An especially simple linear combination isIn terms of the displayed basis, the same vector is $-c_1/7+11c_2/7$. The column space itself is unique; the freedom is in choosing its basis.

No. Take the one-by-one matrices $A=[1]$, $B=[-1]$. Both column spaces are $C=\mathbb{R}$, whereas $A+B=[0]$ has column space $\{0\}$. More generally, taking $B=-A$ cancels any nonzero column space. One always has $\mathrm{Col}(A+B)\subseteq C$, because each sum column lies in the vector subspace $C$, but equality can fail through cancellation.

For an $m\times n$ matrix, its row space is the span of its rows, a subspace of ${\mathbb{R}}^n$; its column space is the span of its columns, a subspace of ${\mathbb{R}}^m$. Its null space is $\ker A=\{x\in {\mathbb{R}}^n:Ax=0\}$. The rank of a matrix is the dimension of its column space, equal to the dimension of its row space. The rank-nullity theorem gives $\dim \ker A=n-\mathrm{rank}A$.

The pivot columns are columns 1, 2 and 3 of the original matrix. A basis for its column space is thereforeUse the original columns, rather than the columns of the row-reduced matrix: elementary row operations preserve column dependence relations, but can change the column space as a subspace of ${\mathbb{R}}^5$.

Writing the columns as $a_1,a_2,a_3$, matrix multiplication givesThe column space is the linear span of these columns, so this particular linear combination belongs to it. The assertion holds for every vector of coefficients $y$.

If $Ax=b$ has a solution, the multiplication identity from part (c) expresses $b$ as a linear combination of the columns, so $b\in \mathrm{Col}(A)$.

Conversely, if $b$ lies in that column space, its definition supplies scalars $x_1,x_2,x_3$ with $b=x_1{a}_1+x_2{a}_2+x_3{a}_3$. Put those scalars into the column vector $x$. Then $Ax=b$. This proves both directions of the claimed equivalence.

Over the underlying field $F$, the row space is the linear span of the row vectors of $A$, a vector subspace of $F^n$. The column space is the linear span of its column vectors, a vector subspace of $F^m$. The null space isa vector subspace of the domain. The column space is the image of a linear map represented by $A$; the null space is its kernel.

The pivot columns of the reduced matrix in part (i) are columns one and two. Select those columns of the original matrix, giving the column-space basisThey are independent, for example their first two entries have determinant $-2\ne 0$. The remaining original column is the first plus twice the second. Taking the reduced matrix's columns instead would generally give the wrong column space.

The rank of $A$ is the dimension of its column space, equivalently the dimension of its row space or the number of pivot columns in a row-echelon form. For an $n\times n$ matrix it lies between zero and $n$.

Prove both inclusions. If $v$ belongs to the column space of $I-P$, then $v=(I-P)u$ for some $u$ and $Pv=(P-P^2)u=0$. Thus $v\in \mathcal{N}(P)$.

Conversely, if $Pv=0$, then $v=(I-P)v$, which explicitly writes $v$ as an image of a linear map under $I-P$. Hence $v\in \mathcal{C}(I-P)$. The two inclusions prove $\mathcal{N}(P)=\mathcal{C}(I-P)$. No orthogonality assumption was used.

Let $c_1=(1,1,1{)}^T$ and $c_2=(2,-2,6{)}^T$. The remaining columns are $3c_1$ and $2c_2$, so these two vectors span the column space. They are linearly independent: the determinant of their first two coordinates is $-4\ne 0$. Therefore $\{(1,1,1{)}^T,(2,-2,6{)}^T\}$ is a basis of the range, and the matrix rank is two.

Equivalently the range is the plane $y_3=2y_1-y_2$: both displayed basis vectors satisfy this equation, and their two-dimensional span fills that plane.