Past exam of the mathematical methods I course of the University of Oxford / 2019 / 11

For a finite matrix, the series defining the matrix exponential converges absolutely in any matrix norm, because $\Vert A^m/m!\Vert \le \Vert A{\Vert }^m/m!$. Thus the Hermitian adjoint may be taken term by term. Since $A^{†}=A$,Therefore the matrix exponential of a Hermitian matrix is Hermitian.

If $A=\mathrm{diag}(a_1,\dots ,a_n)$ is a diagonal matrix, then $A^m=\mathrm{diag}(a_1^m,\dots ,a_n^m)$. Summing each entry separately givesFor the Hermitian matrix case each $a_i$ is real. The diagonal entry formula itself remains valid for complex diagonal entries.

The upper-triangular matrix has distinct eigenvalues $2,-1$, with respective eigenvectors $(1,0{)}^T$ and $(-2,1{)}^T$. ThusThe defining power series shows that the matrix exponential respects similarity. Applying solution to the diagonal factor givesAs a direct series check, $(A^m{)}_{12}=2(2^m-(-1{)}^m)$ for $m\ge 0$; summing this entry divided by $m!$ gives the same off-diagonal expression.

If $M$ is an $r\times s$ matrix, the product $M^2=MM$ is defined in ordinary matrix multiplication only when $s=r$. Hence the defining equation of an idempotent matrix already requires equal row and column counts:Equivalently, an idempotent linear map is an endomorphism, so the same vector space is both its domain and codomain.

If the idempotent matrix $M$ is invertible, multiply $M^2=M$ on the left by its matrix inverse:Thus a nonidentity idempotent matrix must be singular; the identity itself is both invertible and idempotent.

Using $M^2=M$,Thus $I-M$ is another idempotent matrix. In fact $M(I-M)=(I-M)M=0$: as linear projections, they select the complementary summands in the decomposition $x=Mx+(I-M)x$.

Take determinants of the idempotence equation and use multiplicativity of the determinant:Thus $d=\det M$ satisfies $d(d-1)=0$, provingIf the determinant is one, $M$ is invertible and solution gives $M=I$. Every other idempotent matrix has zero determinant.

For real orthonormal column vectors, $u^{T}u=v^{T}v=1$ and $u^{T}v=v^{T}u=0$. Expand using associativity of matrix multiplication:HenceMore precisely, it is the orthogonal projection onto the span of $u,v$: it fixes both vectors and annihilates every vector orthogonal to both.

The transpose convention presumes real vectors, or the explicitly stated bilinear identities above. For complex vectors orthonormal in the Hermitian inner product, the correct orthogonal projection is $u{u}^{†}+v{v}^{†}$. The printed transpose formula need not work in that setting: $u=(i,0{)}^T$ and $v=(0,1{)}^T$ are Hermitian-orthonormal, but $u{u}^T+v{v}^T=\mathrm{diag}(-1,1)$ has square $I_2$, which is different from itself.

Induction from $M^2=M$ gives $M^j=M$ for every integer $j\ge 1$. Separate the zeroth term of the matrix exponential series:Therefore the exponential of an idempotent linear map isThe last form shows its action directly: it multiplies the image of a linear map of $M$ by $e^{\alpha }$ and fixes its kernel of a linear map. No Hermiticity assumption on the idempotent matrix is needed.

Write $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. The two off-diagonal entries of $M^2=M$ giveor $b(\mathrm{tr}M-1)=c(\mathrm{tr}M-1)=0$. If both $b,c$ vanish, $M$ is diagonal; if either is nonzero, the trace must be one. Thus the trace criterion for a two-by-two idempotent isFor the diagonal case, idempotence further forces $a,d\in \{0,1\}$. The stated alternative is only necessary: $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has trace one but its square is $\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$, so it is not idempotent. Among two-by-two matrices of trace one, the additional condition $\det M=0$ is sufficient by the Cayley-Hamilton theorem, which gives $M^2-M+(\det M)I=0$.