The document discusses linear independence and change of basis in vector spaces. It provides the following key points: 1) Two vectors u and v are linearly dependent if one is a multiple of the other, and independent otherwise. 2) Three vectors u, v, and w are linearly dependent if their coefficients in a linear combination equal 0, and independent otherwise. 3) A change of basis matrix P describes the transformation between two bases {e} and {f} of a vector space, where each vector in {f} is written as a linear combination of the vectors in {e}. The inverse of P transforms vectors back from {f} to {e}.

1. NPTEL – Physics – Mathematical Physics - 1
Lecture 9
Linear independence
Determine whether u and v are linearly independent
i) 𝑢 = (1, 2), 𝑣 = (3, −5), (ii) 𝑢 = (1, −3), 𝑣 = (−2𝑢)
2 vectors are said to be linearly dependent if one is multiple of another.
a)
b)
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u and v are independent
u and v are dependent for 𝑣 = −2𝑢
 Determine whether 3 vectors
𝑢 = (1, 1, 2), 𝑣 = (2, 3, 1) and 𝑤 = (4, 5, 5) are linearly independent.
1 2 4 0
𝑥 [1] + 𝑦 [3] + 𝑧 [5] = [0]
2 1 5 0
If this set has a non-zero solution for (x, y, z) then they are linearly dependent.
The students should check this.
Change of basis
Let {𝑒1, 𝑒2 … … … 𝑒𝑛} is a basis of a vector space v and {𝑓1 … … … … 𝑓𝑛} is another
basis. Suppose there is a relation that exists between the two bases such that,
𝑓1 = 𝑎11𝑒1 + 𝑎12𝑒2 + … … … 𝑎1𝑛𝑒𝑛
𝑓2 = 𝑎21𝑒1 + 𝑎22𝑒2 + … … … 𝑎2𝑛𝑒𝑛
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𝑓𝑛 = 𝑎𝑛1𝑒1 + 𝑎𝑛2𝑒2 + … … … 𝑎𝑛𝑛𝑒𝑛
Then the transpose, P of the above matrix of coefficients is called the basis
matrix.
Theorem 1
Let P be a basis matrix from a basis {e;} to a basis {𝑓𝑗 } and Q be the change of
basis matrix from the basis {𝑓𝑗 } to {𝑒𝑖} back. Then P is invertible and 𝑄 = 𝑃−1
Proof 𝑓𝑖 = ∑𝑛 𝑎𝑖𝑗 𝑒
𝑗 =1
𝑗
(1)
(2)
Also 𝑒𝑖 = ∑𝑛
𝑘=1 𝑏𝑗 𝑘 𝑓
𝑘
Substituting (2) in (1)
𝑓𝑖 = ∑𝑛 𝑎𝑖𝑗(∑𝑛 𝑏𝑗𝑘𝑓𝑘 ) = ∑𝑛 (∑𝑛 𝑎 𝑏 ) 𝑓
𝑗 =1 𝑘=1 𝑘=1 𝑗 =1 𝑖 𝑗 𝑗 𝑘 𝑘
Now, ∑𝑗 𝑎𝑖𝑗𝑏𝑗𝑘 = 𝛿𝑖𝑘

2. NPTEL – Physics – Mathematical Physics - 1
Where 𝛿𝑖𝑘 is the Kronecker delta function with the following properties,
𝛿𝑖𝑘 = 1 for 𝑖 = 𝑘
= 0 for i≠ 𝑘 So, 𝐶𝑖𝑘 = 𝛿𝑖𝑘 so 𝑃𝑄 = 1 Or 𝑃 = 𝑄−1 (proved)
Example
Consider the following bases in 𝑅2.
𝑆1 = {𝑢1 = (1, −2), 𝑢2 = (3, −4)}
𝑆2 = {𝑣1 = (1,3), 𝑣2 = (3,8)}
(i) Find the components of an arbitrary vector (𝑏) in 𝑅2 in basis 𝑆1 = {𝑢1, 𝑢2}.
(ii) Write the change of basis matrix P from 𝑆1to 𝑆2. To do this we have to write
𝑣1 and 𝑣2 in terms of 𝑢1 and 𝑢2.
𝑎
Solution
(𝑏) = 𝑥 ( ) + 𝑦 ( ) ⇒ 𝑥 + 3𝑦 = 𝑎 and −2𝑥 − 4𝑦 = 𝑏
𝑎 1
−2 −4
3
3 1
Thus, (𝑎1𝑏)𝑠1 = (−2𝑎
− 2
𝑏) 𝑢1 + (𝑎 + 2
𝑏)
𝑢2
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