Linear Algebra: The Unifying Theorem

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Questions and Answers

What is the defining characteristic of the Unifying Theorem regarding the transformation T?

  • T is onto
  • T is one-to-one
  • S is linearly independent
  • All of the above (correct)

T is __________.

onto

T is __________.

one-to-one

S is __________.

<p>linearly independent</p>
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S spans __________.

<p>R^n</p>
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S is a __________ for R^n.

<p>basis</p>
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A is __________.

<p>invertible</p>
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Ax=b has a __________ solution.

<p>unique</p>
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Ker(T) = null(A) = __________.

<p>{0}</p>
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Nullity(A) = __________.

<p>0</p>
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Col(A) = range(A) = __________.

<p>R^n</p>
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Row(A) = __________.

<p>R^n</p>
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Rank(A) = __________.

<p>n</p>
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Det(A) != __________.

<p>0</p>
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λ = __________ is not an eigenvalue of A.

<p>0</p>
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Study Notes

The Unifying Theorem

  • Let S be a set of n vectors in R^n, and A be a matrix with transformation T: R^n -> R^n defined by T(x) = Ax.
  • Transformation T is both onto (surjective) and one-to-one (injective), indicating a bijective relationship.
  • Set S spans R^n, confirming that every vector in R^n can be expressed as a linear combination of vectors in S.
  • Set S is linearly independent, establishing that no vector in S can be formed from others in the set.
  • Set S acts as a basis for R^n, fulfilling both spanning and independence criteria.
  • The equation Ax = b has a unique solution, affirming that every b in R^n corresponds to exactly one x.
  • Matrix A is invertible, which signifies that it has an inverse A^(-1).
  • The kernel of T (ker(T)) equals the null space of A (null(A)), and is {0}, confirming that the only solution to Ax = 0 is the trivial solution.
  • Nullity of A is 0, indicating there are no free variables in the system.
  • The column space (col(A)) equals the range of A, which is R^n, confirming full rank.
  • The row space (row(A)) is also R^n, indicating the rows of A span the space.
  • The rank of matrix A is n, meaning it has full rank and all n rows/columns are linearly independent.
  • The determinant of A is non-zero (det(A) != 0), confirming that the matrix is full rank and invertible.
  • Eigenvalue condition: λ = 0 is not an eigenvalue of matrix A, indicating no solutions for Ax = 0 beyond the trivial solution.

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