It is, of course, possible to generalize the considerations of the preceding section to multilinear forms and multiple tensor products. Instead of entering into that part of multilinear algebra, we proceed in a different direction; we go directly after determinants.

Suppose that \(A\) is a linear transformation on an \(n\) -dimensional vector space \(\mathcal{V}\) and let \(w\) be an alternating \(n\) -linear form on \(\mathcal{V}\) . If we write \(\overline{A} w\) for the function defined by \[(\overline{A} w)(x_{1}, \ldots, x_{n}) = w(A x_{1}, \ldots, A x_{n}),\] then \(\overline{A} w\) is an alternating \(n\) -linear form on \(\mathcal{V}\) , and, in fact, \(\overline{A}\) is a linear transformation on the space of such forms. Since (see Section: Alternating forms of maximal degree ) that space is one-dimensional, it follows that \(\overline{A}\) is equal to multiplication by an appropriate scalar. In other words, there exists a scalar \(\delta\) such that \(\overline{A} w=\delta w\) for every alternating \(n\) -linear form \(w\) . By this somewhat roundabout procedure (from \(A\) to \(\overline{A}\) to \(\delta\) ) we have associated a uniquely determined scalar \(\delta\) with every linear transformation \(A\) on \(\mathcal{V}\) ; we call \(\delta\) the determinant of \(A\) , and we write \(\delta=\det A\) . Observe that \(\det\) is neither a scalar nor a transformation, but a function that associates a scalar with each linear transformation.

Our immediate purpose is to study the function \(\det\) . We begin by finding the determinants of the simplest linear transformations, that is, the multiplications by scalars. If \(A x=\alpha x\) for every \(x\) in \(\mathcal{V}\) , then \begin{align} (\overline{A} w)(x_{1}, \ldots, x_{n}) &= w(\alpha x_{1}, \ldots, \alpha x_{n})\\ &= \alpha^{n} w(x_{1}, \ldots, x_{n}) \end{align} for every alternating \(n\) -linear form \(w\) ; it follows that \(\det A=\alpha^{n}\) . We note, in particular, that \(\operatorname{det} 0=0\) and \(\operatorname{det} 1=1\) .

Next we ask about the multiplicative properties of \(\det\) . Suppose that \(A\) and \(B\) are linear transformations on \(\mathcal{V}\) , and write \(C=A B\) . If \(w\) is an alternating \(n\) -linear form, then \begin{align} (\overline{C} w)(x_{1}, \ldots, x_{n}) &= w(A B x_{1}, \ldots, A B x_{n})\\ &= (\overline{A} w)(B x_{1}, \ldots, B x_{n})\\ &= (\overline{B} \overline{A} w)(x_{1}, \ldots, x_{n}), \end{align} so that \(\overline{C}=\overline{B} \overline{A}\) . Since \[\overline{C}_{w}=(\operatorname{det} C) w\] and \[\overline{B} \overline{A} w=(\operatorname{det} B) \overline{A} w=(\operatorname{det} B)(\operatorname{det} A) w,\] it follows that \[\operatorname{det}(A B)=(\operatorname{det} A)(\operatorname{det} B).\] (The values of \(\det\) are scalars, and therefore commute with each other.)

A linear transformation \(A\) is called singular if \(\operatorname{det} A=0\) and non-singular otherwise. Our next result is that \(A\) is invertible if and only if it is non-singular. Indeed, if \(A\) is invertible, then \[1=\operatorname{det} 1=\operatorname{det}(A A^{-1})=(\operatorname{det} A)(\operatorname{det} A^{-1}),\] and therefore \(\operatorname{det} A \neq 0\) . Suppose, on the other hand, that \(\operatorname{det} A \neq 0\) . If \(\{x_{1}, \ldots, x_{n}\}\) is a basis in \(\mathcal{V}\) , and if \(w\) is a non-zero alternating \(n\) -linear form on \(\mathcal{V}\) , then \((\operatorname{det} A) w(x_{1}, \ldots, x_{n}) \neq 0\) by Section: Alternating forms , Theorem 3. This implies, by Section: Alternating forms , Theorem 2, that the set \(\{A x_{1}, \ldots, A x_{n}\}\) is linearly independent (and therefore a basis); from this, in turn, we infer that \(A\) is invertible.

In the classical literature determinant is defined as a function of matrices (not linear transformations); we are now in a position to make contact with that approach. We shall derive an expression for \(\operatorname{det} A\) in terms of the elements \(\alpha_{i j}\) of the matrix corresponding to \(A\) in some coordinate system \(\{x_{1}, \ldots, x_{n}\}\) . Let \(w\) be a non-zero alternating \(n\) -linear form; we know that \[(\operatorname{det} A) w(x_{1}, \ldots, x_{n})=w(A x_{1}, \ldots, A x_{n}). \tag{1}\] If we replace each \(A x_{j}\) in the right side of (1) by \(\sum_{i} \alpha_{i j} x_{i}\) and expand the result by multilinearity, we obtain a long linear combination of terms such as \(w(z_{1}, \ldots, z_{n})\) , where each \(z\) is one of the \(x\) ’s. (Compare this part of the argument with the proof of Section: Alternating forms , Theorem 3.) If, in such a term, two of the \(z\) ’s coincide, then, since \(w\) is alternating, that term must vanish. If, on the other hand, all the \(z\) ’s are distinct, then \(w(z_{1}, \ldots, z_{n})=\pi w(x_{1}, \ldots, x_{n})\) for some permutation \(\pi\) , and, moreover, every permutation \(\pi\) can occur in this way. The coefficient of the term \(\pi w(x_{1}, \ldots, x_{n})\) is the product \(\alpha_{\pi(1), 1} \ldots \alpha_{\pi(n), n}\) . Since ( Section: Alternating forms , Theorem 1) \(w\) is skew symmetric, it follows that \[\operatorname{det} A=\sum_{\pi}(\operatorname{sgn} \pi) \alpha_{\pi(1), 1} \ldots \alpha_{\pi(n), n} \tag{2}\] where the summation is extended over all permutations \(\pi\) in \(\mathcal{S}_{n}\) . (Recall that \(w(x_{1}, \ldots, x_{n}) \neq 0\) , by Section: Alternating forms , Theorem 3, so that division by \(w(x_{1}, \ldots, x_{n})\) is legitimate.)

From this classical equation (2) we could derive many special properties of determinants by straightforward computation. Here is one example. If \(\sigma\) and \(\pi\) are permutations (in \(\mathcal{S}_{n}\) ), then (since \(\pi \sigma\) is also a permutation), it follows that the products \(\alpha_{\pi(1), 1} \ldots \alpha_{\pi(n), n}\) and \(\alpha_{\pi \sigma(1), \sigma(1)} \ldots \alpha_{\pi \sigma(n), \sigma(n)}\) differ in the order of their factors only. If, for each \(\pi\) , we take \(\sigma=\pi^{-1}\) , and then alter each summand in (2) accordingly, we obtain \[\operatorname{det} A=\sum_{\pi}(\operatorname{sgn} \pi) \alpha_{1, \pi(1)} \ldots \alpha_{n, \pi(n)}.\] (Note that \(\operatorname{sgn} \pi=\operatorname{sgn} \pi^{-1}\) and that the sum over all \(\pi\) is the same as the sum over all \(\pi^{-1}\) .) Since this last sum is just like the sum in (2), except that \(\alpha_{i, \pi(i)}\) appears in place of \(\alpha_{\pi(i), i}\) , it follows from an application of (2) to \(A^{\prime}\) in place of \(A\) that \[\operatorname{det} A=\operatorname{det} A^{\prime}.\] 

Here is another useful fact about determinants. If \(\mathcal{M}\) is a subspace invariant under \(A\) , if \(B\) is the transformation \(A\) considered on \(\mathcal{M}\) only, and if \(C\) is the quotient transformation \(A / \mathcal{M}\) , then \(\operatorname{det} A=\operatorname{det} B \cdot \operatorname{det} C\) . This multiplicative relation holds if, in particular, \(A\) is the direct sum of two transformations \(B\) and \(C\) . The proof can be based directly on the definition of determinants, or, alternatively, on the expansion obtained in the preceding paragraph.

If, for a fixed linear transformation \(A\) , we write \(p(\lambda)=\operatorname{det}(A-\lambda)\) , then \(p\) is a function of the scalar \(\lambda\) ; we assert that it is, in fact, a polynomial of degree \(n\) in \(\lambda\) , and that the coefficient of \(\lambda^{n}\) is \((-1)^{n}\) . For the proof we may use the notation of (1). It is easy to see that \(w((A-\lambda) x_{1}, \ldots, (A-\lambda) x_{n})\) is a sum of terms such as \(\lambda^{k} w(y_{1}, \ldots, y_{n})\) , where \(y_{i}=x_{i}\) for exactly \(k\) values of \(i\) and \(y_{i}=A x_{i}\) for the remaining \(n-k\) values of \(i\) ( \(k=0,1, \ldots, n\) ). The polynomial \(p\) is called the characteristic polynomial of \(A\) ; the equation \(p=0\) , that is, \(\operatorname{det}(A-\lambda)=0\) , is the characteristic equation of \(A\) . The roots of the characteristic equation of \(A\) (that is, the scalars \(\alpha\) such that \(\operatorname{det}(A-\alpha)=0\) ) are called the characteristic roots of \(A\) .

EXERCISES