
\section{Elementary Matrices and Determinants II}

In the last section, we saw the definition of the determinant and derived an elementary matrix that exchanges two rows of a matrix.  Next, we need to find elementary matrices corresponding to the other two row operations; multiplying a row by a scalar, and adding a multiple of one row to another.  As a consequence, we will derive some important properties of the determinant.

Consider $M=\colvec{R^1 \\ \vdots \\ R^n }$, where $R^i$ are row vectors.  Let $R^i(\lambda)$ be the identity matrix, with the $i$th diagonal entry replaced by $\lambda$, not to be confused with the row vectors. {\it I.e.}
$$
R^i(\lambda)=
\begin{pmatrix}
1 & & & & \\
  & \ddots & & & \\
  & & \lambda & & \\
  & & & \ddots & \\
  & & & & 1 \\
\end{pmatrix}
\, .$$
Then:

\[
M'=R^i(\lambda)M=\colvec{R^1 \\ \vdots \\ \lambda R^i \\ \vdots \\ R^n }
\]
What effect does multiplication by $R^i(\lambda)$ have on the determinant?

\begin{eqnarray*}
\det M' & = & \sum_{\sigma} \text{sgn}(\sigma) m^1_{\sigma(1)}\ldots \lambda m^i_{\sigma(i)} \ldots m^n_{\sigma(n)} \\
& = & \lambda \sum_{\sigma} \text{sgn}(\sigma) m^1_{\sigma(1)}\ldots m^i_{\sigma(i)} \ldots m^n_{\sigma(n)} \\
& = & \lambda \det M
\end{eqnarray*}
Thus, multiplying a row by $\lambda$ multiplies the determinant by $\lambda$.
{\it I.e.} $$\det R^i(\lambda) M = \lambda \det M\, .$$
Since $R^i(\lambda)$ is just the identity matrix with a single row multiplied by $\lambda$, then by the above rule, the determinant of $R^i(\lambda)$ is $\lambda$.  Thus:

\[
\det R^i(\lambda) = \det \begin{pmatrix}
1 & & & & \\
  & \ddots & & & \\
  & & \lambda & & \\
  & & & \ddots & \\
  & & & & 1 \\
\end{pmatrix} = \lambda
\]

The final row operation is adding $\lambda R^j$ to $R^i$.  This is done with the matrix~$S^i_j(\lambda)$, which is an identity matrix but with a $\lambda$ in the $i,j$ position.

\[
S^i_j(\lambda) = \begin{pmatrix}
1 & 	& 	& 	& & & 	\\
  & \ddots & 	&	& & &	\\
  & 	& 1 	& 	& \lambda & &	\\
  & 	& 	& \ddots & & &	\\
  & 	& 	& 	& 1 & & 	\\
  & 	& 	& 	& 	& \ddots & 	\\
  & 	& 	& 	& 	& 	 & 1	\\
\end{pmatrix}
\]
Then multiplying $S^i_j(\lambda)$ by $M$ gives the following:

\[
\begin{pmatrix}
1 & 	& 	& 	& & & 	\\
  & \ddots & 	&	& & &	\\
  & 	& 1 	& 	& \lambda & &	\\
  & 	& 	& \ddots & & &	\\
  & 	& 	& 	& 1 & & 	\\
  & 	& 	& 	& 	& \ddots & 	\\
  & 	& 	& 	& 	& 	 & 1	\\
\end{pmatrix}\colvec{\\ \vdots \\ R^i \\ \vdots \\ R^j \\ \vdots\\ \\}
=
\colvec{\\ \vdots \\ R^i +\lambda R^j \\ \vdots \\ R^j \\ \vdots\\ \\ }
\]
What is the effect of multiplying by $S^i_j(\lambda)$ on the determinant?  Let $M'=S^i_j(\lambda)M$, and let $M''$ be the matrix $M$ but with $R^i$ replaced by $R^j$.

\begin{eqnarray*}
\det M' & = & \sum_{\sigma} \text{sgn}(\sigma) m^1_{\sigma(1)}\ldots (m^i_{\sigma(i)}+ \lambda m^j_{\sigma(j)}) \ldots m^n_{\sigma(n)} \\
& = & \sum_{\sigma} \text{sgn}(\sigma) m^1_{\sigma(1)}\ldots m^i_{\sigma(i)} \ldots m^n_{\sigma(n)} \\
&   & \qquad + \sum_{\sigma} \text{sgn}(\sigma) m^1_{\sigma(1)}\ldots \lambda m^j_{\sigma(j)} \ldots m^j_{\sigma(j)} \ldots m^n_{\sigma(n)} \\
& = & \det M + \lambda \det M''
\end{eqnarray*}
Since $M''$ has two identical rows, its determinant is $0$.  Then $$\det S^i_j(\lambda)M = \det M\, .$$
Notice that if $M$ is the identity matrix, then we have $$\det S^i_j(\lambda) = \det (S^i_j(\lambda)I) = \det I = 1\, .$$
We now have an elementary matrices associated to each of the row operations.

\[
\begin{array}{cccc}
E^i_j &=& I \text{ with rows $i,j$ swapped;} &\det E^i_j=-1 \\[3mm]
R^i(\lambda) &=& I \text{ with $\lambda$ in position $i,i$;} 
	&\det R^i_j(\lambda)=\lambda \\[3mm]
S^i_j(\lambda) &=& I \text{ with $\lambda$ in position $i,j$;} 
	&\det S^i_j(\lambda)=1 \\[3mm]
\end{array}
\]
We have also proved the following theorem along the way:

\begin{theorem}
If $E$ is \emph{any} of the elementary matrices $E^i_j, R^i(\lambda), S^i_j(\lambda)$, then $\det(EM)=\det E \det M$.
\end{theorem}

We have seen that any matrix $M$ can be put into reduced row echelon form via a sequence of row operations, and we have seen that any row operation can be emulated with left matrix multiplication by an elementary matrix.  Suppose that $\rref(M)$ is the reduced row echelon form of $M$.  Then $\rref(M)=E_1E_2\ldots E_kM$ where each $E_i$ is an elementary matrix.

What is the determinant of a square matrix in reduced row echelon form?  
\begin{itemize}
\item If $M$ is not invertible, then some row of $\rref(M)$ contains only zeros.  Then we can multiply the zero row by any constant $\lambda$ without changing~$M$; by our previous observation, this scales the determinant of $M$ by $\lambda$.  Thus, if $M$ is not invertible, $\det \rref(M)=\lambda \det \rref(M)$, and so $\det \rref(M)=0$.  

\item Otherwise, every row of $\rref(M)$ has a pivot on the diagonal; since $M$ is square, this means that $\rref(M)$ is the identity matrix.  Then if $M$ is invertible, $\det \rref(M)=1$.

\item Additionally, notice that $\det \rref(M) = \det (E_1E_2\ldots E_kM)$.  Then by the theorem above, $\det \rref(M)=\det (E_1) \ldots \det (E_k) \det M$.  Since each $E_i$ has non-zero determinant, then $\det \rref(M)=0$ if and only if $\det M=0$.
\end{itemize}
Then we have shown:

\begin{theorem}
For any square matrix $M$, $\det M\neq 0$ if and only if $M$ is invertible.
\end{theorem}
Since we know the determinants of the elementary matrices, we can immediately obtain the following:

\begin{corollary}
Any elementary matrix $E^i_j, R^i(\lambda), S^i_j(\lambda)$ is invertible, except for $R^i(0)$.  In fact, the inverse of an elementary matrix is another elementary matrix.
\end{corollary}


To obtain one last important result, suppose that $M$ and $N$ are square $n\times n$ matrices, with reduced row echelon forms such that, for elementary matrices  $E_i$ and $F_i$, $$M=E_1E_2\ldots E_k \, \rref(M)\, ,$$ and  $$N=F_1F_2\ldots F_l \, \rref(N)=N\, .$$  If $\rref(M)$ is the identity matrix (ie, $M$ is invertible), then:

\begin{eqnarray*}
\det (MN) & = & \det (E_1E_2\ldots E_k\,  \rref(M) F_1F_2\ldots F_l \, \rref(N) )\\
& = & \det (E_1E_2\ldots E_k I F_1F_2\ldots F_l\,  \rref(N) )\\
& = & \det (E_1) \ldots \det(E_k)\det(I)\det(F_1)\ldots\det(F_l)\det(\rref(N)\\
& = & \det(M)\det(N)
\end{eqnarray*}
Otherwise, $M$ is not invertible, and $\det M=0=\det 
\rref(M)$.  Then there exists a row of zeros in $
\rref(M)$, so $R^n(\lambda)
\rref(M)=
\rref(M)$.  Then:
\begin{eqnarray*}
\det (MN) & = & \det (E_1E_2\ldots E_k 
\, \rref(M) N )\\
& = & \det (E_1E_2\ldots E_k 
\, \rref(M) N )\\
& = & \det (E_1) \ldots \det(E_k)\det( 
\rref(M)N)\\
& = & \det (E_1) \ldots \det(E_k)\det( R^n(\lambda) 
\, \rref(M)N)\\
& = & \det (E_1) \ldots \det(E_k)\lambda \det( 
\rref(M)N)\\
& = & \lambda \det (MN)
\end{eqnarray*}
Which implies that $\det (MN)=0=\det M \det N$.

Thus we have shown that for {\it any} matrices $M$ and $N$, 
\[
\det (MN) = \det M \det N
\]
This result is {\it extremely important}; do not forget it!

\section*{References}
Hefferon, Chapter Four, Section I.1 and I.3

Wikipedia:
\begin{itemize}
\item \href{http://en.wikipedia.org/wiki/Determinant}{Determinant}
\item \href{http://en.wikipedia.org/wiki/Elementary_matrix}{Elementary Matrix}
\end{itemize}



\section*{Review Questions}

\begin{enumerate}
\item Let $M=\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$ and $N=\begin{pmatrix}
x & y \\
z & w
\end{pmatrix}$.  Compute the following:
\begin{itemize}
\item $\det M$.
\item $\det N$.
\item $\det (MN)$.
\item $\det M \det N$.
\item $\det (M^{-1})$ assuming $ab-cd\neq 0$.
\item $\det (M^T)$
\item $\det (M+N) - \det M - \det N$
\end{itemize}

\item Suppose $M=\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$ is invertible.  Write $M$ as a product of elementary row matrices times $\rref(M)$.

\item Find the inverses of each of the elementary matrices, $E^i_j, R^i(\lambda), S^i_j(\lambda)$.  Make sure to show that the elementary matrix times its inverse is actually the identity.

\end{enumerate}

\newpage



