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I understand that the ridge regression estimate is the $\beta$ that minimizes residual sum of square and a penalty on the size of $\beta$

$$\beta_{ridge} = (\lambda I_D + X'X)^{-1}X'y = \text{argmin}\: RSS + \lambda ||\beta||^2_2$$

However, I don't fully understand the significance of the fact that $\beta_{ridge}$ differs from $\beta_{OLS}$ by only adding a small constant to the diagonal of $X'X$. Indeed,

$$\beta_{OLS} = (X'X)^{-1}X'y$$

1. My book mentions that this makes the estimate more stable numerically -- why?

2. Is numerical stability related to the shrinkage towards 0 of the ridge estimate, or it's just a coincidence?

I understand that the ridge regression estimate is the $\beta$ that minimizes residual sum of square and a penalty on the size of $\beta$

$$\beta_\mathrm{ridge} = (\lambda I_D + X'X)^{-1}X'y = \operatorname{argmin}\big[ \text{RSS} + \lambda \|\beta\|^2_2\big]$$

However, I don't fully understand the significance of the fact that $\beta_\text{ridge}$ differs from $\beta_\text{OLS}$ by only adding a small constant to the diagonal of $X'X$. Indeed,

$$\beta_\text{OLS} = (X'X)^{-1}X'y$$

1. My book mentions that this makes the estimate more stable numerically -- why?

2. Is numerical stability related to the shrinkage towards 0 of the ridge estimate, or it's just a coincidence?

I understand that the ridge regression estimate is the $\beta$ that minimizes residual sum of square and a penalty on the size of $\beta$

$$\beta_{ridge} = (\lambda I_D + X'X)^{-1}X'y = argmin RSS + \lambda ||\beta||^2_2$$

However, I don't fully understand the significance of the fact that $\beta_{ridge}$ differs from $\beta_{OLS}$ by only adding a small constant to the diagonal of $X'X$. Indeed,

$$\beta_{OLS} = (X'X)^{-1}X'y$$

1. My book mentions that this makes the estimate more stable numerically -- why?

2. Is numerical stability related to the shrinkage towards 0 of the ridge estimate, or it's just a coincidence?

I understand that the ridge regression estimate is the $\beta$ that minimizes residual sum of square and a penalty on the size of $\beta$

$$\beta_{ridge} = (\lambda I_D + X'X)^{-1}X'y = \text{argmin}\: RSS + \lambda ||\beta||^2_2$$

However, I don't fully understand the significance of the fact that $\beta_{ridge}$ differs from $\beta_{OLS}$ by only adding a small constant to the diagonal of $X'X$. Indeed,

$$\beta_{OLS} = (X'X)^{-1}X'y$$

1. My book mentions that this makes the estimate more stable numerically -- why?

2. Is numerical stability related to the shrinkage towards 0 of the ridge estimate, or it's just a coincidence?

I understand that the ridge regression estimate is the $\beta$ that minimizes residual sum of square and a penalty on the size of $\beta$

$$\beta_{ridge} = (\lambda I_D + X'X)^{-1}X'y = argmin RSS + \lambda ||\beta||^2_2$$

However, I don't fully understand the significance of the fact that $\beta_{ridge}$ differs from $\beta_{OLS}$ by only adding a small constant to the diagonal of $X'X$. Indeed,

$$\beta_{OLS} = (X'X)^{-1}X'y$$

1. My book mentions that this makes the estimate more stable numerically -- why?

2. Is numerical stability related to the sparsity of the ridge estimate, or it's just a coincidence?

I understand that the ridge regression estimate is the $\beta$ that minimizes residual sum of square and a penalty on the size of $\beta$

$$\beta_{ridge} = (\lambda I_D + X'X)^{-1}X'y = argmin RSS + \lambda ||\beta||^2_2$$

However, I don't fully understand the significance of the fact that $\beta_{ridge}$ differs from $\beta_{OLS}$ by only adding a small constant to the diagonal of $X'X$. Indeed,

$$\beta_{OLS} = (X'X)^{-1}X'y$$

1. My book mentions that this makes the estimate more stable numerically -- why?

2. Is numerical stability related to the shrinkage towards 0 of the ridge estimate, or it's just a coincidence?