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The shapes in $M{M}^T$ are $(3\times n)(n\times 3)$, whose inner dimensions always match. Hence every positive integer $n$ is allowed, and the resulting matrix is $3\times 3$.

The product of a $3\times n$ matrix with another $3\times n$ matrix requires its inner dimensions $n$ and three to match. Thus $\boxed{n=3}$ is necessary and sufficient for $M^2$ to be defined.

Use Gaussian elimination. The last equation gives $y=c-z$, and the second then gives $x=b-4c-2z$. Substitute these into the first equation to obtain $3b-10c-z=a$. Back-substitution yieldsThe linear system has matrix formThe coefficients of the arbitrary right-hand side in the solved expressions are the rows of the matrix inverse. ThereforeMultiplying the original matrix by this candidate gives the identity matrix; its determinant is $-1$, so the solution is unique for every $(a,b,c)$.

A homogeneous linear system has a nonzero solution exactly when its coefficient matrix is singular. Expansion of the determinant givesThus the allowed values are $\boxed{\alpha =1\text{ or }-8/5}$. Solving the first two equations and checking the third givesIn each case $z\ne 0$ gives a nonzero solution and spans the one-dimensional null space; $z=0$ gives only the zero vector.

Because $\mathbf{b}\cdot \mathbf{x}$ is a scalar, applying the linear operator again givesThis is the first nontrivial case of the powers of a rank-one operator; its matrix is the outer product $\mathbf{a}{\mathbf{b}}^T$.

True. Applying the matrices in order gives $AB\mathbf{x}=A(b\mathbf{x})=ab\mathbf{x}$. The eigenvalue isThis conclusion only uses the shared eigenvector; it does not require $A$ and $B$ to commute on other vectors.

If $A\mathbf{v}=\lambda \mathbf{v}$ for a nonzero eigenvector, then $A^2\mathbf{v}={\lambda }^2\mathbf{v}=A\mathbf{v}=\lambda \mathbf{v}$. Thus $\lambda (\lambda -1)=0$, soBoth values need not occur: the zero matrix has only $0$, and the identity matrix only $1$. More precisely, an idempotent linear map whose rank of a linear map is $r$ has $r$ copies of $1$ and $n-r$ copies of $0$. Indeed, its image of a linear map and kernel of a linear map form a direct sum, giving an eigenbasis without assuming an orthogonal projection.

Take the diagonal matrix $T=\mathrm{diag}(1,0,0)$. Multiplying on the left retains the first row of $S$, while multiplying on the right retains its first column. ThusTheir $(1,3)$ entries differ, so this is an explicit example of noncommuting matrices.

Apply the preceding trace identity first to the pair $AB,C$ and then to $CA,B$:This is invariance under cyclic permutation. It does not assert invariance under arbitrary rearrangement of the matrices.

Expanding the trace of the product and exchanging the two finite summation indices givesScalar entries commute even though the matrices need not.

The Leibniz formula for determinants gives $\det (A^T)={\sum }_{\sigma \in S_3}\mathrm{sgn}(\sigma ){\prod }_i{a}_{\sigma (i),i}$. Relabel the product by $j=\sigma (i)$ and then the sum by $\tau ={\sigma }^{-1}$. The inverse permutation has the same sign, soThe proof applies to any finite square matrix.

The coefficient matrix has determinantA homogeneous linear system has a nonzero solution exactly when this is a singular matrix. ThereforeAt this value all solutions are $(x,y,z)=t(1,-2,1)$, with $t\ne 0$ for a nonzero solution.

Let $N=\sqrt{8}A$ be the unscaled matrix. Expansion along its third row givesScaling a three-by-three matrix by $1/\sqrt{8}$ scales its determinant by $(\sqrt{8}{)}^{-3}$. Therefore

The inverse appearing in the assumption requires $\det C\ne 0$. Taking determinants of $(BC{)}^{\mathsf{T}}=-C^{-1}B$ giveswhere the minus sign comes from the odd matrix order. Multiplying by $\det C$ gives $\det B((\det C{)}^2+1)=0$. Since $C$ is real, the second factor is positive. Thus $\boxed{\det B=0}$ and $B$ is a singular matrix. This is the singularity forced by an odd-dimensional transpose-inverse identity, with the roles of the two matrix letters exchanged.

The characteristic polynomial isIts roots give the two eigenvaluesTheir sum is zero, agreeing with the trace of the matrix.

For the two-by-two matrix, take the product of the diagonal entries minus the product of the off-diagonal entries:In particular, $A$ is an invertible matrix.

For $\lambda =5$, the eigenvalue equation reduces to a one-dimensional eigenspace spanned by $(1,2,-1)$. For $\lambda =-3$, the rows of $A+3I$ are multiples of $(1,2,-3)$, so its eigenspace is the plane $x+2y-3z=0$. One possible set of three independent eigenvectors isEvery eigenvector for $-3$ is a nonzero linear combination of the last two. Their two-dimensional eigenspace shows that the repeated root is not defective: the matrix is diagonalizable.

The characteristic polynomial, using the convention $\det (\lambda I-A)$, isIts zeros give the eigenvalues, including algebraic multiplicity:Their product is $45$ and their sum is $-1$, agreeing with the determinant and trace of the matrix.

For the particular eigenvectors chosen in (iii), the dot products areThus only ${\mathbf{e}}_1$ and ${\mathbf{e}}_2$ are orthogonal for this choice. The matrix is not symmetric, so distinct eigenvalues alone do not imply orthogonality. Another basis within the repeated eigenspace may change which chosen vectors are perpendicular; the dot products above specify the convention completely.

Take determinants of $(CB{)}^{\mathsf{T}}=-B^{-1}C$. In order three, multiplication of a matrix by minus one changes its determinant's sign. HenceThe assumption $\det (B^{\mathsf{T}})\ne 0$ is the same as $\det B\ne 0$, so multiplication by $\det B$ givesSince the matrices are real, the first factor is strictly positive. Thus $\boxed{\det C=0}$, proving that $C$ is a singular matrix. This is singularity forced by an odd-dimensional transpose-inverse identity; no orthogonality of $B$ was used.