By Andrew Baker

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**Example text**

5 1 42 3. LINEAR TRANSFORMATIONS Solution. For the row rank we have 1 2 1 2 1 2 1 0 A = 0 3 ∼ 0 3 ∼ 0 1 ∼ 0 1 = A , 5 1 0 −9 0 −9 0 0 so the row rank is 2. For the column rank, AT = which gives 1 0 5 1 0 5 1 0 5 ∼ ∼ =A , 2 3 1 0 3 −9 0 1 −3 1 0 (A )T = 0 1 5 −3 and the column rank is 2. 40. Let A ∈ Mm×n (F ). Then row rank of A = column rank of A. 40. 10. Recall the linear transformation fA : F n −→ F m determined by fA (x) = Ax (x ∈ F n ). First we identify the value of the row rank of A in other terms.

N of A occurring with suitable multiplicities. 2) λ1 + · · · + λn = tr A = −cn−1 , λ1 · · · λn = det A = (−1)n c0 . 58 5. 3. Eigenspaces and multiplicity of eigenvalues In this section we will work over the field of complex numbers C. 8. Let A be an n × n complex matrix and let λ ∈ C be an eigenvalue of A. Then the λ-eigenspace of A is EigA (λ) = {v ∈ Cn : Av = λv} = {v ∈ Cn : (λIn − A)v = 0}, the set of all eigenvectors associated with λ together with the zero vector 0. If A is a real matrix and if λ ∈ R is a real eigenvalue, then the real λ-eigenspace of A is n n EigR A (λ) = {v ∈ R : Av = λv} = EigA (λ) ∩ R .

The kernel (or nullspace) of f is the following subset of V : Ker f = {v ∈ V : f (v) = 0} ⊆ V. t. w = f (v)} ⊆ W. 14. Let f : V −→ W be a linear transformation. Then (a) Ker f is a subspace of V , (b) Im f is a subspace of W . Proof. (a) Let s1 , s2 ∈ F and v1 , v2 ∈ Ker f . , that f (s1 v1 + s2 v2 ) = 0. Since f is a linear transformation and f (v1 ) = 0 = f (v2 ), we have f (s1 v1 + s2 v2 ) = s1 f (v1 ) + s2 f (v2 ) = s1 0 + s2 0 = 0, hence f (s1 v1 + s2 v2 ) ∈ Ker f . Therefore Ker f is a subspace of V .