17 17 votes Consider three non-zero matrices $A, ~B$ and $C$ such that $ABB^T = CBB^T$ where $B^T$ is the transpose of $B$. Which of the following statements is necessarily true?$r(A) =r(C)$non-zero eigenvalues of $A$ and $C$ are identical.$AB = CB$None of the above. Linear Algebra goclasses goclasses-cs-dpp goclasses-cs-dpp-day-247 linear-algebra goclasses-linear-algebra-practice-questions rank-of-matrix matrix eigen-value + – GO Classes 713 views answer comment Share Follow Print See all 2 Comments 2 2 Comments reply Raman21 commented Jul 8 reply Follow flag Given in the question that A,B,C are non zero matrices$ABB^{T}$ = $CBB^{T}$ $ABB^{T}$ - $CBB^{T}$ = 0$(A-C)BB^{T}$ = 0as B is non zero so $BB^{T}$ can not be zero so A-C = 0A = Cso AB = BC\[B=\begin{bmatrix}1 & 2 & 3 & 4\\5 & 6 & 7 & 8\\9 & 10 & 11 & 12\end{bmatrix},\qquadB^T=\begin{bmatrix}1 & 5 & 9\\2 & 6 & 10\\3 & 7 & 11\\4 & 8 & 12\end{bmatrix}.\]Now,\[BB^T=\begin{bmatrix}1^2+2^2+3^2+4^2 &1\cdot5+2\cdot6+3\cdot7+4\cdot8 &1\cdot9+2\cdot10+3\cdot11+4\cdot12\\[2mm]5\cdot1+6\cdot2+7\cdot3+8\cdot4 &5^2+6^2+7^2+8^2 &5\cdot9+6\cdot10+7\cdot11+8\cdot12\\[2mm]9\cdot1+10\cdot2+11\cdot3+12\cdot4 &9\cdot5+10\cdot6+11\cdot7+12\cdot8 &9^2+10^2+11^2+12^2\end{bmatrix}=\begin{bmatrix}30 & 70 & 110\\70 & 174 & 278\\110 & 278 & 446\end{bmatrix}.\]Observe that the diagonal entries are\[(BB^T)_{11}=1^2+2^2+3^2+4^2=30,\]\[(BB^T)_{22}=5^2+6^2+7^2+8^2=174,\]\[(BB^T)_{33}=9^2+10^2+11^2+12^2=446.\]Hence, each diagonal entry of \(BB^T\) is the sum of squares of the corresponding row of \(B\). Since \(B\neq 0\), at least one row of \(B\) is nonzero, so at least one diagonal entry of \(BB^T\) is positive. Therefore,\[BB^T \neq 0.\] 0 0 replyShare 20asr03 commented Aug 13 reply Follow flag "B is non zero so BB^T can not be zero so A-C = 0"This is wrong interpretation as multiplication of two non-zero matrices can result in a zero matrix 0 0 replyShare Please log in or register to add a comment.
3 3 votes Let $X=A-C$. Then the given condition becomes$XBB^T=0$.Now multiply on the right by $X'$:$XBB^TX^T=0$.But $XBB^TX^T=(XB)(XB)^T$.So we have$(XB)(XB)^T=0$.A matrix of the form $YY^T$ is zero only when $Y=0$. Hence$XB=0$.That is,$(A-C)B=0$,so$AB=CB$.Therefore option C is necessarily true. GO Classes answered Apr 16 GO Classes comment Share Follow 0 reply Please log in or register to add a comment.