03-12-03 - Lecture 17 Andrei Antonenko 1 Kernel its...

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Unformatted text preview: Lecture 17 Andrei Antonenko March 12, 2003 1 Kernel: its dimension and basis Last lecture we saw that the kernel of a linear function is a vector space. Each vector space has a dimension and basis — this lecture we’ll try to determine them for the kernel. Let f : V → U be a linear function, V and U are vector spaces, dim V = n , and dim U = m . We saw that for any linear function we can determine its matrix. So, for f there exists m × n- matrix A A = a 11 a 12 ... a 1 n a 21 a 22 ... a 2 n . . . . . . . . . . . . . . . . . . . a m 1 a m 2 ... a mn such that for any vector x ∈ V f ( x ) = Ax, i.e. if x is a vector such that x = ( x 1 ,x 2 ,...,x n ), then f ( x ) = Ax = a 11 a 12 ... a 1 n a 21 a 22 ... a 2 n . . . . . . . . . . . . . . . . . . . a m 1 a m 2 ... a mn x 1 x 2 . . . x n By definition, vector x belongs to the kernel of f if and only if f ( x ) = . But since f ( x ) = Ax , then this condition can be written as f ( x ) = 0 ⇐⇒ Ax = 0 ⇐⇒ a 11 a 12 ... a 1 n a 21 a 22 ... a 2 n . . . . . . . . . . . . . . . . . . . a m 1 a m 2 ... a mn x 1 x 2 . . . x n = . . . So, the last matrix equality can be written as a linear system a 11 x 1 + a 12 x 2 + ··· + a 1 n x n = 0 a 21 x 1 + a 22 x 2 + ··· + a 2 n x n = 0 ............................................ a m 1 x 1 + a m 2 x 2 + ··· + a mn x n = 0 1 This is a homogeneous system, and we got that its solution space is the kernel of f . We already know how to get a dimension and basis of the solution set for any homogeneous system. The dimension of it will be equal to the number of free variables in REF, and to find a basis we have to solve it, and get particular solutions by assigning 1 to the first free variable and 0’s...
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This note was uploaded on 05/28/2011 for the course AMS 2010 taught by Professor Andant during the Spring '03 term at SUNY Stony Brook.

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03-12-03 - Lecture 17 Andrei Antonenko 1 Kernel its...

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