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Det(E(R1 )), det(A) are all non-zero. Hence this cannot happen and so A = In , showing that A is invertible. There is an explicit formula, Cramer’s Rule, for finding the inverse of a matrix using determinants. However, it is not much use for computations with large matrices and methods based on row and column operations are usually much more efficient. In practise, evaluation of determinants is often simplified by using elementary row operations to create zero’s or otherwise reduced calculations.

Ii) Let V be a finite dimensional vector space of dimension n and let f : V −→ V be a linear transformation. Then χf (f ) = 0, where the right hand side is the constant function taking value 0. Informally, these results are sometimes summarised by saying that a square matrix A or a linear transformation f : V −→ V satisfies its own characteristic equation. 17. Show that the matrix A = 1 0 0 satisfies a polynomial identity of 0 1 3 degree 3. Hence show that A is invertible and express A−1 as a polynomial in A.

2. LINEAR TRANSFORMATIONS 31 so A = fA (e1 ) · · · fA (en ) = Ae1 · · · Aen . Now it easily follows that fA (e1 ), . . , fA (en ) spans Im fA since every element of the latter is a linear combination of this sequence. 11. Let f : V −→ W and g : U −→ V be two linear transformations between vector spaces over F . Then the composition f ◦ g : U −→ W is also a linear transformation. Proof. For s1 , s2 ∈ F and u1 , u2 ∈ V , we have to show that f ◦ g(s1 u1 + s2 u2 ) = s1 f ◦ g(u1 ) + s2 f ◦ g(u2 ).

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Basic Linear Algebra by Andrew Baker

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