# Finding $B,C$ such that $B\left[\begin{smallmatrix}1&2\\4&8\end{smallmatrix}\right]C=\left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$

State $B,C$, such that $B\begin{bmatrix} 1 & 2 \\ 4 & 8 \end{bmatrix}C=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$

I tried some things, but I ended up with non invertible matrices, so I stopped for the moment.

Is there a quick way to guesswork here? (it's an assignment so, maybe some "nice" values for the entries will do)

I don't even know how I multiply a matrix to get 0-Entries with invertible matrices.

So maybe let's start at that part?

• what can you say about the elements of a $2 \times 2$ invertible matrix? – user394255 Jan 6 '17 at 21:32
• @A.Molendijk let the elements be a,b,c,d. Then $ad-bc\not=0$ – SAJW Jan 6 '17 at 21:33
• nice. Are we allowed to use the fact that they are invertible or do we have to show by means of computation? – user394255 Jan 6 '17 at 21:37
• @A.Molendijk it's explicitly staten those matrices are invertible, so I assume that. – SAJW Jan 6 '17 at 21:39
• You could start with setting $B=\begin{pmatrix} b_1 & b_2 \\ b_3 & b_4 \end{pmatrix}$ and in the same way $C$. Then multiply the left hand side out and compare it with the right side. That would be the first thinge I'd try. – Fritz Jan 6 '17 at 21:42

Hint: row-reduce $\begin{bmatrix} 1 & 2 \\ 4 & 8 \end{bmatrix}$, then column-reduce. Row operations are equivalent to multiplication to the left by special invertible matrices (hence $B$) and column operations are equivalent to multiplication to the right (hence $C$).

More details: subtracting 4 times the first row from the second row is attained by $$\begin{bmatrix} 1 & 0 \\ -4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & 8 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$$

Then subtracting twice the first column from the second column is attained by $$\begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$

Added: This method works for matrices of any size. If $A$ is an $m\times n$ matrix of rank $r$, start with $$\begin{array}{cc} I_m & A \\ & I_n \end{array}$$ and row reduce $A$, performing the same row operations on $I_m$, to obtain $$\begin{array}{cc} B & \text{rref}(A) \\ & I_n \end{array}.$$

Then move the copy of $I_r$ to the top left corner and make zeroes everywhere else using column operations; again, perform the same column operations on $I_n$. When that is done one gets $$\begin{array}{cc} B & \begin{bmatrix} I_r & 0 \\ 0 & 0 \end{bmatrix} \\ & C \end{array}$$ and $$BAC = \begin{bmatrix} I_r & 0 \\ 0 & 0\end{bmatrix} .$$ The matrices $B$ and $C$ are not unique.

• Far more elegant than my approach. – Doug M Jan 6 '17 at 21:51
• This is probably what was intended in this exercise. OP would do well to remember that elementary transformations on rows/columns correspond to multiplication by elementary matrices. This is essential for matrix equivalence (rank), finding matrix inverse and solving linear systems. – Ennar Jan 6 '17 at 21:57
• @DougM The OP is really computing the changes of basis that put the matrix in Smith normal form. – Viktor Vaughn Jan 6 '17 at 22:02

$B,C$ are not unique.

$\begin {bmatrix} 1&2\\4&8 \end{bmatrix} = B^{-1} \begin {bmatrix} 1&0\\0&0 \end{bmatrix}C^{-1}$

Suppose $B^{-1} = \begin {bmatrix} b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix}$ and $C^{-1} = \begin {bmatrix} c_{11}&c_{12}\\c_{21}&c_{22} \end{bmatrix}$

$\begin {bmatrix} 1&2\\4&8 \end{bmatrix} = $$\begin {bmatrix} b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix} \begin {bmatrix} 1&0\\0&0 \end{bmatrix}\begin {bmatrix} c_{11}&c_{12}\\c_{21}&c_{22}\end{bmatrix}\\ \begin {bmatrix} b_{11}&b_{12}\\b_{21}&b_{22} \end{bmatrix} \begin {bmatrix} c_{11}&c_{12}\\0&0 \end{bmatrix} \\ \begin {bmatrix} b_{11}c_{11}&b_{11}c_{12}\\b_{21}c_{11}&b_{21}c_{12} \end{bmatrix} b_{11}c_{11} = 1\\ b_{11}c_{12}= 2\\ c_{12} = 2c_{11}\\ b_{21}c_{11} = 4b_{11}\\ Choose the other elements to make your calculation easy. i.e. \det B^{-1} = 1 B = \begin {bmatrix} 1&0\\-4&1 \end{bmatrix} C = \begin {bmatrix} 1&-2\\0&1 \end{bmatrix} Should work. Completely different approach. Daigonlize \begin{bmatrix} 1&2\\4&8 \end{bmatrix} P^{-1}\begin{bmatrix} 1&2\\4&8 \end{bmatrix}P = \begin{bmatrix} 9&0\\0&0 \end{bmatrix} \frac 1{81}\begin{bmatrix} 1&2\\-4&1\end{bmatrix}\begin{bmatrix} 1&2\\4&8 \end{bmatrix}\begin{bmatrix} 1&-2\\4&1\end{bmatrix} = \begin{bmatrix} 1&0\\0&0 \end{bmatrix} I am surprised that no one has mentioned this:$$ B\pmatrix{1\\ 4}\pmatrix{1&2}C=B\pmatrix{1&2\\ 4&8}C=\pmatrix{1&0\\ 0&0}=\pmatrix{1\\ 0}\pmatrix{1&0}. $$So, it suffices to find two invertible matrices B and C such that B\pmatrix{1\\ 4}=\pmatrix{1\\ 0} and \pmatrix{1&2}C=\pmatrix{1&0}. • interesting! thanks – SAJW Jan 7 '17 at 16:04 The characteristic polynomial of the matrix M=\begin{bmatrix}1&2\\4&8\end{bmatrix} is \enspace\chi_N(\lambda)=\lambda^2-9\lambda. Thus we have the following eigenvalues and eigenvectors:$$\lambda=9:\enspace e_1=\begin{bmatrix}1\\4\end{bmatrix} ,\qquad\lambda=0:\enspace e_2=\begin{bmatrix}2\\-1\end{bmatrix}.$$In the basis (e_1, e_2), with change of basis matrix P=\begin{bmatrix}1&2\\4&-1\end{bmatrix}, the matrix becomes$$P^{-1}MP=\begin{bmatrix}9&0\\0&0\end{bmatrix}.$$Now pre-(or post-)multiply this matrix by the invertible A=\begin{bmatrix}\frac19&0\\0&1\end{bmatrix}. You obtain the relation$$(AP^{-1})MP=\begin{bmatrix}1&0\\0&0\end{bmatrix}$$so that a solution is B=AP^{-1}, \;C=P. Another is BP^{-1}, \;C=PA. We have that$$\begin {bmatrix} 1&2\\4&8 \end{bmatrix} = P \begin {bmatrix} 9&0\\0&0 \end{bmatrix}P^{-1} = (3P) \begin {bmatrix} 1&0\\0&0 \end{bmatrix} (3P^{-1}) $$where P \in \mathrm{Mat}_2(\mathbb{Q}). Thus,$$B\begin{bmatrix} 1 & 2 \\ 4 & 8 \end{bmatrix}C=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \Longleftrightarrow (3BP)\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}( 3P^{-1}C) =\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$or,$$(3BP)\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} =\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} (3P^{-1}C)^{-1}.$$Then we obtain that$$3BP = \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} ; (3P^{-1}C)^{-1} = \begin{bmatrix} a & 0 \\ d & e \end{bmatrix}$$where b, d \in \mathbb{C} and a,c,e \in \mathbb{C}^{*} (because B, C are invertible). Then$$B = \frac{1}{3} \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} P^{-1} ; \quad C= \frac{1}{3} P \begin{bmatrix} a & 0 \\ d & e \end{bmatrix}^{-1}; a,c,e \in \mathbb{C}^*; b, d \in \mathbb{C}.$\$