Mir­ror images and undo­ing

If a square matrix is unchanged when you trans­pose it, we call it sym­met­ric. If trans­pos­ing just flips all its signs, it is skew-sym­met­ric. The nice thing is that every square matrix can be split into one of each. Some square matri­ces can also be undone, that is, they have inverses. We will prove the split­ting, say exactly what an invert­ible matrix is, find inverses using ele­men­tary row oper­a­tions, and end with mixed prac­tice.

Sym­met­ric and skew-sym­met­ric matri­ces

AA is sym­met­ric if AT=AA^T = A, skew-sym­met­ric if AT=−AA^T = -A. Notice that a skew-sym­met­ric matrix must have zeros on its diag­o­nal (aii=−aiia_{ii} = -a_{ii}).

Now take any square AA. Then A+ATA + A^T is sym­met­ric and A−ATA - A^T is skew-sym­met­ric, so

A=12(A+AT)+12(A−AT).\displaystyle A = \tfrac{1}{2}(A + A^T) + \tfrac{1}{2}(A - A^T).

Exam­ple 1. A=[43−12]A = \begin{bmatrix} 4 & 3 \\ -1 & 2 \end{bmatrix}. Here P=12(A+AT)=[4112]\displaystyle P = \tfrac{1}{2}(A + A^T) = \begin{bmatrix} 4 & 1 \\ 1 & 2 \end{bmatrix}, Q=12(A−AT)=[02−20]\displaystyle Q = \tfrac{1}{2}(A - A^T) = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}, and indeed P+Q=AP + Q = A.

Exam­ple 2. If AA and BB are sym­met­ric of the same order, ABAB is sym­met­ric exactly when AB=BAAB = BA, because (AB)T=BTAT=BA(AB)^T = B^TA^T = BA.

Invert­ible matri­ces

A square matrix AA is invert­ible if there is BB with AB=BA=IAB = BA = I. Then we write B=A−1B = A^{-1}, and there is only one such matrix. (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}.

There are three ele­men­tary row oper­a­tions: swap two rows, mul­ti­ply a row by a non-zero num­ber, and add a mul­ti­ple of one row to another. To find A−1A^{-1}, write A=IAA = IA and apply row oper­a­tions to AA on the left and to II on the right until the left side becomes II. What­ever stands on the right is then A−1A^{-1}. If a row of zeros turns up on the left, stop: AA has no inverse.

Exam­ple 3. A=[2513]A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}. [25∣1013∣01]\left[\begin{smallmatrix} 2 & 5 & | & 1 & 0 \\ 1 & 3 & | & 0 & 1 \end{smallmatrix}\right]; R1↔R2R_1 \leftrightarrow R_2: [13∣0125∣10]\left[\begin{smallmatrix} 1 & 3 & | & 0 & 1 \\ 2 & 5 & | & 1 & 0 \end{smallmatrix}\right]; R2→R2−2R1R_2 \to R_2 - 2R_1: [13∣010−1∣1−2]\left[\begin{smallmatrix} 1 & 3 & | & 0 & 1 \\ 0 & -1 & | & 1 & -2 \end{smallmatrix}\right]; R2→−R2R_2 \to -R_2, then R1→R1−3R2R_1 \to R_1 - 3R_2: A−1=[3−5−12]A^{-1} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}. Always check your answer: AA−1=IAA^{-1} = I.

Exam­ple 4. [2436]\begin{bmatrix} 2 & 4 \\ 3 & 6 \end{bmatrix}: R2→R2−32R1\displaystyle R_2 \to R_2 - \tfrac{3}{2}R_1 gives a zero row, so there is no inverse.

Mixed prac­tice

  1. Show that A=[03−2−3052−50]A = \begin{bmatrix} 0 & 3 & -2 \\ -3 & 0 & 5 \\ 2 & -5 & 0 \end{bmatrix} is skew-sym­met­ric.
  2. Write [351−1]\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} as the sum of a sym­met­ric and a skew-sym­met­ric matrix.
  3. Write [14−20362−25]\begin{bmatrix} 1 & 4 & -2 \\ 0 & 3 & 6 \\ 2 & -2 & 5 \end{bmatrix} as a sum of sym­met­ric and skew-sym­met­ric parts.
  4. Show that for any square AA, AATAA^T is sym­met­ric.
  5. Find the inverse by row oper­a­tions: [3152]\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}; [4−31−1]\begin{bmatrix} 4 & -3 \\ 1 & -1 \end{bmatrix}.
  6. Show that [6−3−21]\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} has no inverse.
  7. Find A−1A^{-1} for A=[120013001]A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{bmatrix}.
  8. If A=[cos⁡α−sin⁡αsin⁡αcos⁡α]A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}, show ATA=IA^TA = I and hence A−1=ATA^{-1} = A^T.
  9. If AA and BB are invert­ible, ver­ify (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} by mul­ti­ply­ing.
  10. Find kk if A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} sat­is­fies A2−5A−kI=OA^2 - 5A - kI = O.

Answers to check against

Show answers
  1. AT=−AA^T = -A entry by entry.
  2. [333−1]+[02−20]\begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}.
  3. P=[120232025]P = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 3 & 2 \\ 0 & 2 & 5 \end{bmatrix}, Q=[02−2−2042−40]Q = \begin{bmatrix} 0 & 2 & -2 \\ -2 & 0 & 4 \\ 2 & -4 & 0 \end{bmatrix}.
  4. (AAT)T=(AT)TAT=AAT(AA^T)^T = (A^T)^TA^T = AA^T.
  5. [2−1−53]\begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}; [1−31−4]\begin{bmatrix} 1 & -3 \\ 1 & -4 \end{bmatrix}.
  6. R1→R1+3R2R_1 \to R_1 + 3R_2 gives a zero row.
  7. [1−2601−3001]\begin{bmatrix} 1 & -2 & 6 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{bmatrix}.
  8. Mul­ti­ply­ing out, you get cos⁡2α+sin⁡2α=1\cos^2\alpha + \sin^2\alpha = 1 on the diag­o­nal, 00 off it.
  9. (AB)(B−1A−1)=A(BB−1)A−1=AA−1=I(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AA^{-1} = I.
  10. A2=[7101522]A^2 = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}; A2−5A=[2002]A^2 - 5A = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}: k=2k = 2.