I have some trouble understanding the property that says the value of a determinant won't depend on which row or column we choose. My textbook says that if we rearrange the terms after expanding the cofactors it will lead to this conclusion, but is there any other intuitive way to think of this? Also, why the determinant value reaches zero when there is two identical rows or columns in the matrix we choose to calculate its determinant?
Asked
Active
Viewed 227 times
2 Answers
3
Determinants can be very non-intuitive, perhaps the best visualisation of them in 3D is of a signed volume. Write $[v_1, v_2, v_3]$ for the matrix with column vectors $v_1$, $v_2$, and $v_3$ respectively. You may know that $| \det [v_1 v_2 v_3]|$ is the volume of the Parallelepiped spanned by those vectors. The determinant $\det [v_1 v_2 v_3]$ (without the absolute value signs) is a signed volume, meaning it remembers something about the ordering of the vectors, and may be negative instead of positive.
With this interpretation, we have some properties:
- If $v_1 = v_2$, the parallelepiped is degenerate and has no volume - it's just a parallelogram. So $\det [v_1 v_1 v_3] = 0$.
- Any time two vectors are interchanged, the signed volume switches sign, but stays the same magnitude. So $\det [v_1 v_2 v_3] = - \det [v_3 v_2 v_1] = \det [v_2 v_3 v_1]$ and so on.
- If one of the vectors doubles in length, the parallelepiped doubles in volume. $\det [(\lambda v_1) v_2 v_3] = \lambda \det[v_1 v_2 v_3]$ for any scalar $\lambda$.
By the second rule there, you can rearrange the vectors however you like, and you'll get the same determinant up to a sign change. Hopefully this explains a little about why it doesn't matter which column you expand down.
Joppy
- 12,875
1
This might not be the direct intuition you are looking for, but here is one approach to be aware of that I think seems pretty clear.
Key Idea: Often the best way to think of the function $A \mapsto \det A$ is that it is a multilinear, alternating function of the columns of $A$ which satisfies $\det I = 1$ (and the determinant is the only function with these three properties).
If we take these three properties as a starting point, we can understand the properties of the determinant that you mentioned pretty easily. First of all, "alternating" means that if we interchange two columns of $A$, then the sign of the determinant changes. But, if $A$ has two identical columns, then clearly interchanging those two columns will not change the value of the determinant. How is it possible that the sign of the determinant flips and yet the determinant does not change value? This is only possible if $\det A = 0$.
Having established this fact, it can easily be shown that adding a scalar multiple of one column to another column does not change the value of the determinant. ($\spadesuit$)
We can now derive the "expansion by minors" formula, as follows.
For simplicity, assume that
$$
A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33} \end{bmatrix}.
$$
Notice that
$$
\begin{bmatrix}
a_{11}\\a_{21}\\a_{31}
\end{bmatrix}
= a_{11} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}
+a_{21} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}
+a_{31} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.
$$
The fact that the determinant is multilinear tells us that
\begin{align}
\det A &= a_{11} \det \begin{bmatrix} 1 & a_{12} & a_{13} \\
0 & a_{22} & a_{23} \\
0 & a_{32} & a_{33}
\end{bmatrix} +
a_{21} \det \begin{bmatrix} 0 & a_{12} & a_{13} \\
1 & a_{22} & a_{23} \\
0 & a_{32} & a_{33}
\end{bmatrix} +
a_{31} \det \begin{bmatrix} 0 & a_{12} & a_{13} \\
0 & a_{22} & a_{23} \\
1 & a_{32} & a_{33}
\end{bmatrix}\\
&= a_{11} \det \begin{bmatrix} 1 & 0 & 0 \\
0 & a_{22} & a_{23} \\
0 & a_{32} & a_{33}
\end{bmatrix} +
a_{21} \det \begin{bmatrix} 0 & a_{12} & a_{13} \\
1 & 0 & 0 \\
0 & a_{32} & a_{33}
\end{bmatrix} +
a_{31} \det \begin{bmatrix} 0 & a_{12} & a_{13} \\
0 & a_{22} & a_{23} \\
1 & 0 & 0
\end{bmatrix}\\
\end{align}
In the last step, I used the rule ($\spadesuit$) mentioned above to introduce some zeros.
Now, notice that the function $$ \begin{bmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{bmatrix} \mapsto \det \begin{bmatrix} 1 & 0 & 0 \\ 0 & a_{22} & a_{23} \\ 0 & a_{32} & a_{33} \end{bmatrix} $$ satisfies the three key properties that it is multilinear, alternating, and maps the identity matrix to $1$. It follows that $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & a_{22} & a_{23} \\ 0 & a_{32} & a_{33} \end{bmatrix} = \det \begin{bmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{bmatrix}. $$ Similar reasoning allows us to conclude that $$ \det A = a_{11} \det \begin{bmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{bmatrix} - a_{21} \det \begin{bmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{bmatrix} + a_{31} \det \begin{bmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{bmatrix}. $$ This is the "expansion by minors" formula when we expand along the first column of $A$. With a similar argument we can derive the formula for expanding along any other column of $A$, and you can see what the formula will be.
littleO
- 51,938