Past exam of the mathematics course of the University of Oxford / 2015 / Prelims

Source: released examination PDF, bundle pages 1–4.

For the linear map $T:V\to W$, its image of a linear map and kernel of a linear map areBoth contain the zero vector, since $T(0)=0$. If $y_1=T(v_1)$ and $y_2=T(v_2)$ belong to the image of a linear map, then for any real numbers $a,b$,Thus $\mathrm{im}T$ is a vector subspace of $W$. If $u_1,u_2\in \ker T$, thenso $\ker T$ is a vector subspace of $V$.

The rank of a linear map and nullity of a linear map areThe rank-nullity theorem states thatTo prove it, choose a basis $u_1,\dots ,u_k$ of the kernel of a linear map and use basis extension to obtain a basisof $V$. The vectors $T(v_1),\dots ,T(v_{n-k})$ span the image of a linear map: applying $T$ to an arbitrary linear combination of the displayed basis removes all the $u_i$ terms. They are also linearly independent. Indeed,so ${\sum }_j{a}_j{v}_j={\sum }_i{b}_i{u}_i$ for some scalars $b_i$. The linear independence of the extended basis forces every $a_j$ and $b_i$ to vanish. Consequently $r(T)=n-k$ and $n(T)=k$, proving the identity. The same argument covers an empty basis when one of these dimensions is zero.

Here $\dim V=2$. By the rank-nullity theorem, $r(T)+n(T)=2$, and a dimension is a nonnegative integer. All three possibilities occur:The images of a linear map in these examples have dimensions $0,1,2$, respectively, while the corresponding kernel of a linear map is ${\mathbb{R}}^2$, $\{(0,y):y\in \mathbb{R}\}$ and $\{0\}$. The identity above excludes all other pairs.

If $TS={\mathrm{id}}_W$, every $w\in W$ would satisfy $w=T(Sw)$, so $T$ would be a surjective linear map. But its image of a linear map has dimension at most $\dim V=2$, whereas $\dim W=3$. Therefore no such $S$ exists. This is the obstruction to a right inverse: $T$ cannot reach every target vector.

The necessary and sufficient condition isFor necessity, if $ST={\mathrm{id}}_V$ and $v\in \ker T$, then $v=S(Tv)=S(0)=0$.

For sufficiency, let $e_1,e_2$ be a basis of $V$ and suppose the kernel of a linear map is trivial. The vectors $T{e}_1,T{e}_2$ are linearly independent, since $aT{e}_1+bT{e}_2=0$ implies $a{e}_1+b{e}_2\in \ker T$. Extend them to a basis $T{e}_1,T{e}_2,w_3$ of $W$. Define a linear map $S$ byThen $ST$ agrees with the identity operator on a basis, hence on all of $V$. Thus a left inverse exists precisely when $T$ is injective; this is the linear left inverse criterion.

An elementary row operation is one of three reversible changes: interchange two rows; multiply one row by a nonzero scalar; or add a scalar multiple of one row to a different row. Their inverses are, respectively, the same interchange, multiplication by the reciprocal, and addition of the negative scalar multiple.

Applied to an augmented matrix, these changes interchange linear equations, rescale an equation by a nonzero factor, or add a multiple of one equation to another. Every old solution satisfies the new equations. Applying the inverse elementary row operation gives the converse, so the solution set is unchanged at every step. Equivalently, all the operations together givewhere $E$ is an invertible matrix, a product of elementary matrices.

Writing the unknown vector as $(x,y,z{)}^T$, the first two linear equations give $x=z-1$ and $y=z-\beta$. Substitution into the third givesThus, when $\alpha \ne 2$, put $t=-\beta /(\alpha -2)$. There is exactly one solution:When $\alpha =2$, the remaining equation is $0=-\beta$. HenceThe second case is an affine line in a vector space, so it contains infinitely many solutions. These cases exhaust all real numbers $\alpha ,\beta$.

A system of linear equations $T_{\alpha }x=b$ is solvable exactly when $b\in \mathrm{im}{T}_{\alpha }$. If $x_0$ is one solution, thenso its solution set is $x_0+\ker T_{\alpha }$, a translate of the kernel of a linear map.

For $\alpha \ne 2$, the matrix is an invertible matrix, andEvery $b$ is in the image of a linear map and has exactly one preimage. For $\alpha =2$, its first two columns span the image of a linear map, while the corresponding homogeneous linear system gives the kernel of a linear map:The prescribed $b=(-1,\beta ,1{)}^T$ satisfies the displayed condition precisely when $\beta =0$. If $\beta \ne 0$, it lies outside the image of a linear map and there is no solution. If $\beta =0$, it lies in the image of a linear map, and the nonzero kernel of a linear map supplies infinitely many solutions.

For $\alpha \ne 2$, $A^{\prime }=I_3$, whose row space is ${\mathbb{R}}^3=\mathrm{im}{T}_{\alpha }$. For $\alpha =2$, elementary row operations give the reduced row echelon formIts row space is $\{(u,v,w):u+v+w=0\}$, whereas the image of a linear map is $\{(u,v,w):u-v+w=0\}$. These differ: $(0,1,-1)$ belongs to the former but not the latter. Thus (A) is true exactly when $\alpha \ne 2$. Rows are being identified with vectors in ${\mathbb{R}}^3$ in this comparison.

Every elementary column operation preserves the linear span of the columns: each new column is a linear combination of the old columns, and the inverse operation supplies the reverse inclusion. Thus the column space of $A^{\prime \prime }$ equals that of $A$, which is precisely the image of a linear map $T_{\alpha }$. Consequently (B) is true for every $\alpha \in \mathbb{R}$.

For $\alpha \ne 2$, both reduced forms are $I_3$, so $A^{\prime \prime }=(A^{\prime }{)}^T$. At $\alpha =2$, the reduced column echelon form is the transpose of the reduced row echelon form of $A^T$, givingFor example, $C_2\mapsto -C_2$ followed by $C_3\mapsto C_3+C_1-C_2$, using the updated second column, produces the displayed $A^{\prime \prime }$ directly. Its third column is zero. The two matrices differ, so (C) is true exactly when $\alpha \ne 2$. Taking a transpose exchanges the kind of reduction, but also replaces the original $A$ by $A^T$; it does not generally identify the two reduced forms of $A$ itself.

Choose the rotation taking $e_1$ to $e_2$, $e_2$ to $-e_1$ and $e_3$ to itself. The columns of the matrix of a linear map are these images of the standard basis, soThis fixes the orientation used throughout this question.

Using the convention ${\chi }_T(x)=\det (xI-A)$, the characteristic polynomial isSince $x^2+1>0$ for every $x\in \mathbb{R}$, the only eigenvalue over the real numbers is $1$. Solving $(A-I)v=0$ givesThus the only eigenspace over the real numbers is the axis of the rotation.

The two nonzero proper invariant subspaces areBoth are preserved by the rotation. To show there are no others, note thatis the linear projection onto $U$, and $I-P$ is the linear projection onto $W$. Every invariant subspace $L$ is preserved by $I,T^2$, and hence by both of these linear projections. ThereforeBecause $U$ has dimension one, $L\cap U$ is either $0$ or $U$. If $L\cap W$ contains a nonzero $v=(a,b,0)$, it also contains $Tv=(-b,a,0)$. These two vectors are linearly independent, because their two-dimensional determinant is $a^2+b^2>0$. They span $W$, so $L\cap W$ is either $0$ or $W$. The only possibilities for $L$ are therefore $0,U,W,V$.

Finally every $(x,y,z)$ has the unique decomposition $(0,0,z)+(x,y,0)$, with one summand in each of $U,W$. HenceThis also illustrates how invariant subspaces of a quarter-turn in three dimensions can be classified by linear projections.

Over the complex numbers, the characteristic polynomial splits into three distinct roots of a polynomial, $1,i,-i$. Direct multiplication gives the eigenvectorsThey are linearly independent: the last two have two-dimensional determinant $2i\ne 0$ in the first two coordinates, and the first has nonzero third coordinate. ThusThis is a basis of eigenvectors of ${\mathbb{C}}^3$, proving that $T^{\prime }$ has a diagonalizable matrix.

The prescribed form is a sum of products of real polynomial evaluations. It is a symmetric bilinear form, since each evaluation is a linear map. AlsoEquality forces $f(-1)=f(0)=f(1)=0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, by the root bound for a polynomial. Thus equality holds only for the zero polynomial, establishing positive definiteness. Therefore the form is a real inner product, and $V$ is an inner product space. This is a discrete sampling inner product on polynomial spaces.

The relevant inner products areHence $1$ and $x$ are orthogonal vectors. Dividing by their norms gives the orthonormal basisof their linear span $U$.

For any orthonormal basis $f,g$ of $U$, bilinearity givesTogether with $⟨f,g⟩=0$, these identities prove pairwise orthogonality. The construction subtracts the orthogonal projection onto a finite-dimensional subspace $U$ from $x^2$, leaving the orthogonal projection residual.

Use $f,g$ from i. Their inner products with $x^2$ areThus $h=x^2-2/3$. Its values at $-1,0,1$ are $1/3,-2/3,1/3$, soBy ii the three polynomials are orthogonal vectors, and normalizing $h$ givesThis orthonormal set is a basis of the three-dimensional vector space $V$. It is the result of the Gram-Schmidt process applied to $1,x,x^2$.

If $T$ is an orthogonal transformation and $e_1,e_2,e_3$ is an orthonormal basis, thenThe images form an orthonormal set, hence are linearly independent and form an orthonormal basis in dimension three.