Elementary Column Operations of a Matrix. The following operations on a matrix A are called elementry column operations.
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- Interchange of any two columns of A.
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- Multiplications of a column of A by a non zero scalar.
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- Addition of a scalar multiple of one column of A to another column.
On applying an elementary column operation to the identity matrix , we obtain a corresponding column matrix, which is non singular.
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Column equivalent to of a matrix
An [katex] m \times n [/katex] matix B is called column equivalent to an [katex] m \times n [/katex] matrix A if B is obtained by performing a sequence of a finite number of elementary column operations on A. we write [katex]B \stackrel{C}{\sim}[/katex] A to denote B is column equivalent to A.
An elementary column operation on a matrix A becomes an elementary row operatio on the transpose [katex] A^{T} [/katex] of A. THus if [katex]B \stackrel{C}{\sim} A[/katex], then [katex]B^{T} \stackrel{C}{\sim} A^{T}[/katex]. Hence [katex]B^{T} =EA^{T}[/katex] where E is an elementary matrix.
Equivalent matrix
If an [katex] m \times n [/katex] matix B can be obtained from an [katex] m \times n [/katex] matrix A a finite number of elementary row and column operations, then B is equivalent to A.
The row and column equivalences are special cases of the general concepts of of euivalentoe of matrices.
Theorem
An [katex]$m \times n[/katex] matrix B is equivalent to an [katex]m \times n[/katex] matrix A if and only if B=PAQ where P and Q are nonsingular and of orders [katex]m \times m[/katex] and [katex]n \times n[/katex] respectively.
Theorem
Every nonzero [katex] m \times n [/katex] matrix is equivaleat to an [katex] m \times n [/katex] matrix D where
D=\left[\begin{array}{ll}
I_{r} & 0 \\
0 & 0
\end{array}\right]D is called the canonical (or normal form of the matrix A).
Proof. Since [katex]A \neq 0[/katex], there exists a nonzero element a of A. By performing elementary row and column operations on A, we obtain an equivalent matrix with element a is the (1, 1) position. Applying the operation [katex]a^{-1} R_1[/katex], we get 1 in the (1,1) position. By adding suitable multiples of the first row to the remaining rows, every other element of the first columa can be made zero. Having done that we can reduce every element of the first row, except that in the (1,1) position, to a zero. We thus obtain an equivalent matrix of the form
B=\left[\begin{array}{ll}
1 & 0 \\
0 & C
\end{array}\right]Here 0 in the first row is a [katex]1 \times(n-1)[/katex] zero matrix, 0 in the first column is an [katex](m-1) \times 1[/katex] zero matrix and C is an [katex](m-1) \times(n-1)[/katex] matrix. Repeating the process with [katex]C_{\text {i we get }}[/katex] ge the required form. The procedure is explained in the following example.
Example
Reduce the matrix
A=\left[\begin{array}{ccc}1 & 1 & 2 \\ 1 & 2 & 3 \\ 0 & -1 & -1\end{array}\right]to the normal form. Also find the nonsingular matrices P and Q such that PAQ is in the normal form.
Solution. Here we write the given matrix A and the identity matrix [katex]I_3[/katex] (twice) in three columns as below. The same row operations as on A are performed on the first identity matrix and the same column operations as on A are performed on the second identity matrix. Thus
A\quad \quad \quad \quad \quad \quad I_3 \quad \quad \quad \quad \quad \quad I_2
Row operations Column operations
\left[\begin{array}{ccc}
1 & 1 & 2 \\
1 & 2 & 3 \\
0 & -1 & -1
\end{array}\right] \quad\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \quad\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right][katex]R_2-R_1[/katex], we have
\left[\begin{array}{rrr}
1 & 1 & 2 \\
0 & 1 & 1 \\
0 & -1 & -1
\end{array}\right]\left[\begin{array}{rrr}
1 & 0 & 0 \\
-1 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]By [katex]C_2-C_1[/katex] and [katex]C_3-2 C_1[/katex], we get
\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & -1 & -1
\end{array}\right]\left[\begin{array}{rrr}
1 & 0 & 0 \\
-1 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{rrr}
1 & -1 & -2 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]For [katex]R_3+R_2[\katex], we obtain
\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{array}\right]\left[\begin{array}{rrr}
1 & 0 & 0 \\
-1 & 1 & 0 \\
-1 & 1 & 1
\end{array}\right]\left[\begin{array}{rrr}
1 & -1 & -2 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]By [katex]C_3-C_2[\katex], we have
\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{array}\right] \quad\left[\begin{array}{rrr}
1 & 0 & 0 \\
-1 & 1 & 0 \\
-1 & 1 & 1
\end{array}\right] \quad\left[\begin{array}{rrr}
1 & -1 & -1 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{array}\right]
\left[\begin{array}{cc}
I_2 & 0_{2 \times 1} \\
0_{1 \times 2} & 0_{1 \times 1}
\end{array}\right] \quad \quad \quad P \quad \quad \quad \quad QThus the required normal form is
\left[\begin{array}{cc}I_2 & 0 \\ 0 & 0\end{array}\right],with
P=\left[\begin{array}{rrr}
1 & 0 & 0 \\
-1 & 1 & 0 \\
-1 & 1 & 1
\end{array}\right] \text { and } Q=\left[\begin{array}{rrr}
1 & -1 & -1 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{array}\right]Verification:
PAQ=\left[\begin{array}{ccc}
1 & 0 & 0 \\
-1 & 1 & 0 \\
-1 & 1 & 1
\end{array}\right] \quad\left[\begin{array}{lll}
1 & 1 & 2 \\
1 & 2 & 3 \\
0 & -1 & -1
\end{array}\right] \quad\left[\begin{array}{ccc}
1 & -1 & -1 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 1 & 2 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{array}\right] \quad\left[\begin{array}{lll}
1 & -1 & -1 \\
0 & 1 & -1 \\
0 & 0 & 1
\end{array}\right]= \left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{cc}I_2 & 0 \\ 0 & 0\end{array}\right]