No, except when started at the exact solution. Merely failing the usual symmetric positive-definite matrix hypothesis would not by itself prove divergence. For this particular matrix one can prove that every defined nonzero-residual step increases the linear-system residual norm.

Write $r=(x,y{)}^T$, put $D=r^{T}Ar=2x^2-2xy-2y^2$ and $B=5x^2+4xy+3y^2$. For $D\ne 0$, direct substitution into the stated iteration givesFor $r\ne 0$, bothare strictly positive; completing squares gives $7(x+y/7{)}^2+6y^2/7$ and $3(x+y{)}^2+2y^2$. Hence $B>|D|$ and the residual norm strictly increases at every defined step. If $D=0$ before the residual vanishes, the algorithm is instead undefined. Neither behavior converges to residual zero. The matrix is invertible, so this also rules out convergence of the iterates to its unique solution.

For example, from the zero starting guess the first two step sizes are $37/11$ and $-37/11$, giving $r_1=(42,30{)}^T$ and $r_2=(1380/11)r_0$. This demonstrates the failure without treating a sufficient convergence theorem as a necessary one. The general calculation is residual growth in an indefinite steepest-descent iteration.

At $u_0=0$, the linear-system residual is $r_0=b=(1,1{)}^T$. Here $A{r}_0=r_0$, so the steepest-descent step size isConsequently $u_1=(1,1{)}^T$, and $A{u}_1=b$ gives residual zero. The stated exact-arithmetic stopping test with $ϵ=0$ is met after one iteration. Stop there rather than trying to evaluate the next, undefined $0/0$ step. This is the special case where the initial residual is an eigenvector.

The line passes through $p=(0,1{)}^T$ and has the same direction as in part (i). Translate by $-p$, reflect by $H$, then translate back. In homogeneous coordinates the affine reflection isAct on column $(x,y,1{)}^T$, reading the product from right to left. Every point with $y=4x/3+1$ is fixed. The translation column is $p-Hp$, as for any reflection in an affine hyperplane.

A unit vector along the line is $v=(3,4{)}^T/5$. Reflection fixes its component parallel to $v$ and negates its perpendicular component, so the reflection matrix isIndeed $H(3,4{)}^T=(3,4{)}^T$, $H(-4,3{)}^T=-(-4,3{)}^T$ and $H^2=I$. These characterize the orthogonal reflection in that line.

The first column has norm two, giving $q_1=(1,-1,1,-1{)}^T/2$. Its inner products with the second and third columns are both one. Subtracting $q_1$ from the second column leaves $(3,3,3,3{)}^T/2$, of norm three, so $q_2=(1,1,1,1{)}^T/2$. The inner product of $q_2$ with the third column is one. Subtracting $q_1+q_2$ from that column leaves $(1,1,-1,-1{)}^T$, of norm two. Thus the Gram-Schmidt process givesThe columns of $Q$ have unit norm and pairwise zero inner products, so $Q^{T}Q=I_3$. Direct multiplication verifies $QR=A$.

For independent columns $a_1,\dots ,a_k$ in a real inner product space, the Gram-Schmidt process recursively formsEach $u_j$ subtracts the orthogonal projections onto earlier normalized vectors. Their independence ensures $u_j\ne 0$. The resulting $q_i$ are orthonormal and preserve the span of every initial column set. Putting $Q=(q_1\cdots q_k)$ and $r_{ij}=⟨q_i,a_j⟩$ for $i\le j$, with zeros below the diagonal, gives a QR decomposition $A=QR$ with $Q^{T}Q=I$. Dependent input columns require omitting a zero residual or treating rank deficiency separately.

A linear map is one-to-one, or injective, when equal outputs imply equal inputs. Equivalently its kernel is $\{0\}$: subtracting inputs converts equality of outputs to a zero output. Here $T$ is not one-to-one, since its kernel has dimension two. Concretely, $T(-2,-1,1,0)=T(0)=0$ for distinct inputs.

A linear map $T:V\to W$ is onto, or surjective, when every $w\in W$ equals $T(v)$ for some $v\in V$; equivalently its image is all of $W$. Here $T$ is not onto, because its image has dimension two while its declared codomain ${\mathbb{R}}^3$ has dimension three. Indeed every image vector satisfies $x+y-3z=0$, so, for example, $(0,0,1)$ is not attained.

The rank-nullity theorem states $\dim V=\dim \ker T+\dim \mathrm{im}T$ for a linear map from a finite-dimensional vector space $V$. Here the domain has dimension four and the image has dimension two. Thus $\dim \ker T=4-2=2$. For example, a kernel basis is $\{(-2,-1,1,0),(-1,-2,0,1)\}$, as follows from the reduced equations.

Gaussian elimination givesThe first two original columns form an image basis:Their independence follows already from their third and first coordinates. The third column is twice the first plus the second, and the fourth is the first plus twice the second, so they span every output of $T$.

If $B$ is an invertible upper triangular matrix, $B^T$ is invertible and lower triangular. Part (ii) makes $(B^T{)}^{-1}$ lower triangular, while part (i) identifies it with $(B^{-1}{)}^T$. Transposing gives that $B^{-1}$ is upper triangular. This deduction uses both the inverse-transpose identity and the lower-triangular result.

Let column $j$ of $A^{-1}$ be $x$, so $Ax=e_j$. Every diagonal entry of the invertible lower triangular matrix $A$ is nonzero, since its determinant is the product of those entries. In row one through row $j-1$, the right-hand side is zero. Forward substitution first gives $x_1=0$, then $x_2=0$, and successively $x_i=0$ for all $i<j$. These are precisely the entries above the diagonal in column $j$ of $A^{-1}$. Every column satisfies this, proving that $A^{-1}$ is lower triangular. This is triangularity of an inverse matrix.

Transpose both inverse identities, using reversal of multiplication under matrix transposition:Thus $(A^T{)}^{-1}=(A^{-1}{)}^T$, by uniqueness of the matrix inverse.

True. Existence follows because the basis spans $V$. If $v={\sum }_i{a}_i{s}_i={\sum }_i{b}_i{s}_i$, subtract to get ${\sum }_i(a_i-b_i)s_i=0$. The basis vectors' linear independence forces every $a_i-b_i=0$, proving uniqueness. For an infinite basis, both representations have finite support, so the same argument applies to their finite union. These coefficients are the coordinates relative to a basis.

False. In the real vector space $\mathbb{R}$, both $\{1\}$ and $\{2\}$ are bases: each is linearly independent and spans the whole space. A basis need not be a unique set. What is fixed for a finite-dimensional vector space is the number of basis elements, its dimension.

Each elementary matrix is obtained by applying one invertible elementary row operation to the identity matrix. Its inverse is the elementary matrix for the reverse operation. A product of elementary matrices is therefore invertible and has full rank $n$.

Conversely, if $\mathrm{rank}A=n$, Gaussian elimination has a pivot in every column and reduces $A$ to the identity. Writing the successive row operations as left multiplication gives $E_r\cdots E_{1}A=I$. Hence $A=E_1^{-1}\cdots E_r^{-1}$, a product of elementary matrices. This proves both directions of the elementary-matrix factorization theorem.

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 $n$.

The reduced equations are $x+z=0$ and $y+2z=0$. Thus the null space is the lineThe nonzero spanning vector gives a one-element basis, so the nullity is one.

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.

Gaussian elimination givesElementary row operations preserve the row space, so $\{(1,0,1),(0,1,2)\}$ is a basis. The two rows are linearly independent because their first two coordinates form the standard basis of ${\mathbb{R}}^2$.