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Concept 02 / 09Matrices and Determinants

Transpose, Symmetric and Skew-Symmetric Matrices

The transpose and the reversal rule (AB)^T = B^T A^T, symmetric and skew-symmetric matrices, why the diagonal of a skew-symmetric matrix is zero and its determinant of odd order vanishes, and the unique split of any square matrix into symmetric and skew-symmetric parts.

The transpose ATA^{T} (also written A′A') of an m×nm \times n matrix AA is the n×mn \times m matrix obtained by turning rows into columns: (AT)ij=aji(A^{T})_{ij} = a_{ji}. Properties: (AT)T=A(A^{T})^{T} = A, (A+B)T=AT+BT(A + B)^{T} = A^{T} + B^{T}, (kA)T=kAT(kA)^{T} = kA^{T} and, most importantly, (AB)T=BTAT(AB)^{T} = B^{T}A^{T}. The order reverses. To see why, the (i,j)(i, j) entry of (AB)T(AB)^{T} is the (j,i)(j, i) entry of ABAB, which is row jj of AA dotted with column ii of BB, which is row ii of BTB^{T} dotted with column jj of ATA^{T}, the (i,j)(i, j) entry of BTATB^{T}A^{T}. The rule extends to (ABC)T=CTBTAT(ABC)^{T} = C^{T}B^{T}A^{T}, and it is also consistent with sizes: if AA is m×nm \times n and BB is n×pn \times p, then BTATB^{T}A^{T} is (p×n)(n×m)(p \times n)(n \times m), which is defined, while ATBTA^{T}B^{T} usually is not. Symmetric and skew-symmetric. A square matrix is symmetric if AT=AA^{T} = A, that is, aij=ajia_{ij} = a_{ji}: it is a mirror image of itself across the diagonal. It is skew-symmetric if AT=−AA^{T} = -A, that is, aij=−ajia_{ij} = -a_{ji}. Setting i=ji = j in the skew condition gives aii=−aiia_{ii} = -a_{ii}, so every diagonal entry of a skew-symmetric matrix is 00. The only matrix that is both symmetric and skew-symmetric is OO. Closure facts, each a one-line check with the transpose rules: A+ATA + A^{T} is symmetric and A−ATA - A^{T} is skew-symmetric. AATAA^{T} and ATAA^{T}A are always symmetric (for any, even non-square, AA). If A,BA, B are symmetric, ABAB is symmetric if and only if AB=BAAB = BA, since (AB)T=BTAT=BA(AB)^{T} = B^{T}A^{T} = BA. If AA is symmetric, so is BTABB^{T}AB for any BB of suitable order; if AA is skew-symmetric, so is BTABB^{T}AB. Powers: if AA is symmetric, all powers are symmetric; if AA is skew-symmetric, odd powers are skew-symmetric and even powers are symmetric. Decomposition. Every square matrix can be written uniquely as a symmetric matrix plus a skew-symmetric matrix: A=12(A+AT)+12(A−AT)A = \frac{1}{2}(A + A^{T}) + \frac{1}{2}(A - A^{T}). Existence is the formula itself. Uniqueness: if A=P+Q=P′+Q′A = P + Q = P' + Q' with P,P′P, P' symmetric and Q,Q′Q, Q' skew, then P−P′=Q′−QP - P' = Q' - Q is both symmetric and skew-symmetric, hence OO. This is the matrix version of splitting a function into its even and odd parts. Determinant of a skew-symmetric matrix. Since det⁡(AT)=det⁡A\det(A^{T}) = \det A and det⁡(−A)=(−1)ndet⁡A\det(-A) = (-1)^{n}\det A for an n×nn \times n matrix, a skew-symmetric AA satisfies det⁡A=det⁡(AT)=det⁡(−A)=(−1)ndet⁡A\det A = \det(A^{T}) = \det(-A) = (-1)^{n}\det A. If nn is odd, det⁡A=−det⁡A\det A = -\det A, so det⁡A=0\det A = 0. A skew-symmetric matrix of odd order is therefore always singular. For even order the determinant is a perfect square of a polynomial in the entries; for example, det⁡[0a−a0]=a2\det\begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix} = a^{2}. Counting. For n×nn \times n matrices with entries from a set of kk values, a symmetric matrix is determined by the n(n+1)2\frac{n(n+1)}{2} entries on and above the diagonal, giving kn(n+1)/2k^{n(n+1)/2} matrices. A skew-symmetric matrix is determined by the n(n−1)2\frac{n(n-1)}{2} entries above the diagonal (the diagonal is forced to be 00), provided the entry set is closed under negation. Key Takeaways: - (AB)T=BTAT(AB)^{T} = B^{T}A^{T}; (AT)T=A(A^{T})^{T} = A; det⁡(AT)=det⁡A\det(A^{T}) = \det A. - Symmetric: AT=AA^{T} = A. Skew-symmetric: AT=−AA^{T} = -A, with all diagonal entries 00. - A+ATA + A^{T} symmetric; A−ATA - A^{T} skew; AATAA^{T} and ATAA^{T}A symmetric. - Product of symmetric matrices is symmetric iff they commute. - A=12(A+AT)+12(A−AT)A = \frac{1}{2}(A + A^{T}) + \frac{1}{2}(A - A^{T}), uniquely. - A skew-symmetric matrix of odd order has determinant 00. - Symmetric n×nn \times n matrices have n(n+1)2\frac{n(n+1)}{2} free entries; skew-symmetric have n(n−1)2\frac{n(n-1)}{2}. - Pitfall: writing (AB)T=ATBT(AB)^{T} = A^{T}B^{T}. Brain-Teaser 1: AA and BB are symmetric matrices of the same order. Classify AB−BAAB - BA and AB+BAAB + BA. Solution: (AB−BA)T=BTAT−ATBT=BA−AB=−(AB−BA)(AB - BA)^{T} = B^{T}A^{T} - A^{T}B^{T} = BA - AB = -(AB - BA), so AB−BAAB - BA is skew-symmetric. Similarly (AB+BA)T=BA+AB(AB + BA)^{T} = BA + AB, so AB+BAAB + BA is symmetric. As a consequence, the diagonal entries of AB−BAAB - BA are all 00, and for odd order det⁡(AB−BA)=0\det(AB - BA) = 0. Applying the transpose rule and comparing with the original is the universal method for classification questions. Brain-Teaser 2: Express A=[15−12]A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} as the sum of a symmetric and a skew-symmetric matrix. Solution: AT=[1−152]A^{T} = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}. The symmetric part is P=12(A+AT)=12[2444]=[1222]P = \frac{1}{2}(A + A^{T}) = \frac{1}{2}\begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}. The skew part is Q=12(A−AT)=12[06−60]=[03−30]Q = \frac{1}{2}(A - A^{T}) = \frac{1}{2}\begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}. Check: P+Q=AP + Q = A. A quick shortcut: the off-diagonal entries of PP are the averages of the mirror pairs (5+(−1)2=2\frac{5 + (-1)}{2} = 2), and those of QQ are the half-differences (5−(−1)2=3\frac{5 - (-1)}{2} = 3), while the diagonal goes entirely to PP. Brain-Teaser 3: If AA is skew-symmetric and xx is any column vector of the right size, show that xTAx=0x^{T}Ax = 0. Solution: xTAxx^{T}Ax is a 1×11 \times 1 matrix, so it equals its own transpose: xTAx=(xTAx)T=xTAT(xT)T=xTATx=−xTAxx^{T}Ax = (x^{T}Ax)^{T} = x^{T}A^{T}(x^{T})^{T} = x^{T}A^{T}x = -x^{T}Ax. So xTAx=0x^{T}Ax = 0. Written out, xTAx=∑i,jaijxixjx^{T}Ax = \sum_{i,j}a_{ij}x_{i}x_{j}, and the terms aijxixja_{ij}x_{i}x_{j} and ajixjxia_{ji}x_{j}x_{i} cancel in pairs while the diagonal terms are 00. The trick of transposing a scalar is short but powerful, and it is used again for quadratic forms and orthogonal matrices.